Respan Dataset Explorer

David wants to complete a hypothesis test with the least amount of probability for error. If he sets the significance level to 1%, assuming his sample is truly random, what else could he adjust in the test in order to reduce error? 
He could change the population mean.
He could increase the sample size.
He could change the population standard deviation.
He could decrease the sample size.

In hypothesis testing, the significance level (\(\alpha\)) represents the probability of making a Type I error, which occurs when we incorrectly reject the null hypothesis when it is actually true. David is setting his significance level to 1%, which means he wants a very low probability of making this error.

To further reduce error (especially Type II error, which occurs when we fail to reject a false null hypothesis), David can take the following action:

### Options Analysis:

1. **He could change the population mean.**  
   - Changing the population mean does not reduce error within the context of hypothesis testing; instead, it changes the hypothesis being tested.

2. **He could increase the sample size.**  
   - **This is the correct option.** Increasing the sample size results in a larger and more representative sample, which tends to reduce the variability of the sample mean. This, in turn, leads to a more precise estimate of the population parameter, reduces the standard error, and can improve the power of the test (the probability of correctly rejecting a false null hypothesis).

3. **He could change the population standard deviation.**  
   - Altering the population standard deviation is not possible and would not logically apply in the context of reducing error in hypothesis testing. The population standard deviation is a characteristic of the population and cannot be manipulated.

4. **He could decrease the sample size.**  
   - Decreasing the sample size would generally increase variability and reduce the precision of the test, thus potentially increasing both Type I and Type II errors.

### Conclusion:

To reduce error in his hypothesis test, David should **increase the sample size**. This will help improve the accuracy of the results and reduce the likelihood of making errors related to the hypothesis testing process.

Zachary completes a hypothesis test and finds that he rejects the null hypothesis. Which statement gives a reason for rejecting the null hypothesis?
The z-statistic is less than 0.
The z-statistic lies in the critical region.
The z-statistic lies outside the critical region.
The z-statistic is greater than 0.

In hypothesis testing, the decision to reject the null hypothesis is based on the z-statistic and its position relative to critical values. 

### Statement Analysis

1. **The z-statistic is less than 0.**
   - This statement alone does not provide a reason for rejecting the null hypothesis. The sign of the z-statistic (whether it is less than or greater than zero) does not inherently indicate whether it is in the critical region.

2. **The z-statistic lies in the critical region.**
   - This statement is correct. If the z-statistic falls in the critical region (defined by the significance level, \(\alpha\)), this means there is strong evidence against the null hypothesis, leading to its rejection.

3. **The z-statistic lies outside the critical region.**
   - This statement would indicate that we do **not** reject the null hypothesis. If the z-statistic is outside the critical region, we fail to reject the null hypothesis.

4. **The z-statistic is greater than 0.**
   - Like the first statement, the positivity of the z-statistic alone does not determine whether the null hypothesis is rejected. The value of the z-statistic must be considered within the context of the critical values.

### Conclusion

The correct reason for rejecting the null hypothesis is:

**The z-statistic lies in the critical region.** 

This indicates that the evidence is strong enough to conclude that the null hypothesis is not likely true based on the sample data.

A study investigated the job satisfaction of teachers allowed to choose supplementary curriculum for their classes versus teachers who were assigned all curricular resources for use in their classes. The authors of the study wanted to know if the two groups of teachers had different levels of job satisfaction. They will use a significance level of 5% for their test.

Upper-Tail Values
a
5%
2.5%
1%
Critical
z-values	
1.65
1.96
2.58

What is (are) the critical value(s) for z in this study?
–1.65 and 1.65
–1.96 and 1.96
1.65
1.96

In the context of hypothesis testing, when comparing two groups to see if they have different levels of job satisfaction, the authors are interested in determining whether the means of the two groups are statistically different from each other.

Since the study is looking for differences (without specifying a direction), it implies a two-tailed test. In a two-tailed test, we need to consider both tails of the distribution because we are checking for differences in both directions (greater than or less than).

### Given Significance Level and Critical Values

1. **Significance Level (\(\alpha\))**: 5% for a two-tailed test means that the critical regions are split across both tails of the distribution, allocating 2.5% to each tail.

2. The critical z-values at a 5% significance level for a two-tailed test, according to the z-values provided, are:
   - For the upper tail: **1.96**
   - For the lower tail: **-1.96**

### Conclusion

The critical values for \( z \) in this study are:

**–1.96 and 1.96.**

This means if the calculated \( z \)-statistic is less than -1.96 or greater than 1.96, we would reject the null hypothesis, concluding that there is a significant difference in job satisfaction between the two groups of teachers.

A recent survey of 8,000 high school students found that the mean price of a prom dress was $195.00 with a standard deviation of $12.00. Alyssa thinks that her school is more fashion conscious and spent more than $195.00. She collected data from 20 people in her high school and found that the average price spent on a prom dress was $208.00. Which of the following is the correct z-statistic for this situation?
0.24
4.84
7.51
96.90

To calculate the z-statistic for Alyssa's hypothesis test, we can use the formula for the z-statistic when comparing a sample mean to a population mean:

\[
z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
\]

Where:
- \(\overline{x} = 208\) (the sample mean),
- \(\mu = 195\) (the population mean),
- \(\sigma = 12\) (the population standard deviation),
- \(n = 20\) (the sample size).

### Step 1: Calculate the Standard Error (SE)

First, we need to calculate the standard error (SE) of the sample mean:

\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{20}}
\]

Calculating \(\sqrt{20}\):

\[
\sqrt{20} \approx 4.472
\]

Now calculate the SE:

\[
SE = \frac{12}{4.472} \approx 2.684
\]

### Step 2: Calculate the z-statistic

Now we can find the z-statistic:

\[
z = \frac{208 - 195}{2.684} = \frac{13}{2.684} \approx 4.84
\]

### Conclusion

Therefore, the correct z-statistic for this situation is:

**4.84**.

A popular video game claims that the average time needed to reach level 10 Paladin is 3 hours with a standard deviation of 0.4 hours. James thinks that he and his four friends are more skilled than the average gamer because it took them an average of only 2.5 hours. Which of the following is the most restrictive level that would validate his claim?

Upper-Tail Values
a
5%
2.5%
1%
Critical
z-values
1.65
1.96
2.58
1%
2.5%
5%
10%

To determine the most restrictive significance level (α) that would validate James's claim that he and his friends are more skilled than the average gamer, we need to perform a hypothesis test.

### Step 1: Formulate the Hypotheses

1. **Null Hypothesis (\(H_0\))**: The average time to reach level 10 Paladin is equal to 3 hours.  
   \( H_0: \mu = 3 \)

2. **Alternative Hypothesis (\(H_a\))**: The average time to reach level 10 Paladin is less than 3 hours (indicating that they are more skilled).  
   \( H_a: \mu < 3 \)

### Step 2: Calculate the z-statistic

Given:
- Population mean (\(\mu\)) = 3 hours
- Standard deviation (\(\sigma\)) = 0.4 hours
- Sample size (\(n\)) = 5 (James and 4 friends)
- Sample mean (\(\overline{x}\)) = 2.5 hours

First, we calculate the standard error (SE):

\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{0.4}{\sqrt{5}} \approx \frac{0.4}{2.236} \approx 0.1789
\]

Now we calculate the z-statistic:

\[
z = \frac{\overline{x} - \mu}{SE} = \frac{2.5 - 3}{0.1789} \approx \frac{-0.5}{0.1789} \approx -2.79
\]

### Step 3: Determine the Critical z-value

Next, we need to determine the most restrictive level where the z-statistic of approximately -2.79 would indicate significance.

Using the table values provided:
- For \( \alpha = 0.05 \) (5% significance): Critical z-value is approximately -1.645 (for a one-tailed test).
- For \( \alpha = 0.025 \) (2.5% significance): Critical z-value is approximately -1.96.
- For \( \alpha = 0.01 \) (1% significance): Critical z-value is approximately -2.33 (not provided in your options, but known).
- For \( \alpha = 0.005 \) (0.5% significance): Critical z-value is approximately -2.576.

### Step 4: Choose the Most Restrictive Level

For James's claim to be validated with his calculated z-value of approximately -2.79, we need it to be less than the critical z-value of the significance level:

1. **At \(\alpha = 0.05\)**, the critical z-value is approximately -1.645. \(-2.79 < -1.645\) - rejects \(H_0\).
2. **At \(\alpha = 0.025\)**, the critical z-value is approximately -1.96. \(-2.79 < -1.96\) - rejects \(H_0\).
3. **At \(\alpha = 0.01\)**, the critical z-value is -2.33. \(-2.79 < -2.33\) - rejects \(H_0\).
4. **At \(\alpha = 0.005\)**, the z-statistic of -2.79 is not less than the critical z-value of approximately -2.576. **Cannot reject \(H_0\)**.

### Conclusion

Given these calculations and interpretations, the **most restrictive significance level \( \alpha \)** that would validate James's claim while still allowing him to reject the null hypothesis is:

**1%** (since that is the highest level where the z-statistic still leads to rejecting the null hypothesis).

A consumer protection group randomly checks the volume of different beverages to ensure that companies are packaging the stated amount. Each individual volume is not exact, but a volume of iced tea beverages is supposed to average to 300 mL with a standard deviation of 3 mL. The consumer protection group sampled 20 beverages and found the average to be 298.4 mL. Using the given table, which of the following is the most restrictive level of significance on a hypothesis test that would indicate the company is packaging less than the required average 300 mL?

Upper-Tail Values
a
5%
2.5%
1%
Critical
z-values
1.65
1.96
2.58
1%
2.5%
5%
10%

To determine the most restrictive level of significance (\( \alpha \)) that would indicate that the company is packaging less than the required average of 300 mL, we need to form our hypotheses and calculate the z-statistic based on the sample data provided.

### Step 1: Formulate the Hypotheses

1. **Null Hypothesis (\( H_0 \))**: The average volume is 300 mL.  
   \( H_0: \mu = 300\)

2. **Alternative Hypothesis (\( H_a \))**: The average volume is less than 300 mL.  
   \( H_a: \mu < 300\)

### Step 2: Given Information

- Population mean (\( \mu \)) = 300 mL
- Sample mean (\( \overline{x} \)) = 298.4 mL
- Sample size (\( n \)) = 20
- Population standard deviation (\( \sigma \)) = 3 mL

### Step 3: Calculate the Z-Statistic

The formula for the z-statistic is:

\[
z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
\]

First, we calculate the standard error (SE):

\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{20}} \approx \frac{3}{4.472} \approx 0.671
\]

Now plug this into the z-statistic formula:

\[
z = \frac{298.4 - 300}{0.671} = \frac{-1.6}{0.671} \approx -2.38
\]

### Step 4: Determine the Critical Z-Value and Significance Level

Now, we compare the calculated z-value with the critical z-values at various significance levels:

- For \( \alpha = 0.05 \) (5% significance), the critical z-value is \( -1.645 \).
- For \( \alpha = 0.025 \) (2.5% significance), the critical z-value is \( -1.96 \).
- For \( \alpha = 0.01 \) (1% significance), the critical z-value is \( -2.33 \).
- For \( \alpha = 0.005 \) (0.5% significance), the critical z-value is \( -2.576 \).

### Step 5: Compare the Calculated Z-Statistic with Critical Values

- Calculated z-statistic: **-2.38**
  - \( -2.38 \) is less than \( -2.33 \) (1% significance level)
  - \( -2.38 \) is between \( -1.96 \) (2.5% significance level) and \( -2.33 \) (1% significance level)

### Conclusion

Thus, the most restrictive level of significance at which you can conclude that the company is packaging less than the required average of 300 mL is:

**2.5% (with critical z-value of -1.96)**.

You would reject the null hypothesis at this significance level because the z-statistic of -2.38 would fall in the rejection region for \( \alpha = 0.025 \).

Tyesha found that the z-statistic was 2.1 and that the critical z-values were -1.96 and 1.96. Which of the following is a valid conclusion based on these results?
One can reject the null hypothesis.           
One can reject the alternate hypothesis.
One can accept the null hypothesis.     
One cannot accept or reject the null hypothesis.

To draw a conclusion based on Tyesha's findings regarding the z-statistic and the critical z-values, let's examine the details provided:

1. **Calculated z-statistic**: \( z = 2.1 \)
2. **Critical z-values**: \( -1.96 \) (lower tail) and \( 1.96 \) (upper tail)

### Step 1: Understand the Decision Rule

In a two-tailed hypothesis test, you look at both tails of the distribution:

- **Reject the null hypothesis (\(H_0\))**: If the calculated z-statistic falls beyond the critical values in either tail. In this case:
  - If \( z < -1.96 \) or \( z > 1.96 \), reject \( H_0 \).
  
- **Fail to reject the null hypothesis**: If the calculated z-statistic falls within the range defined by the critical values, i.e., \( -1.96 < z < 1.96 \).

### Step 2: Apply the Decision Rule to Tyesha’s Results

- Tyesha's z-statistic is \( 2.1 \).
- Since \( 2.1 > 1.96 \), it falls in the rejection region (beyond the critical value in the upper tail).

### Conclusion

Since the calculated z-statistic (2.1) exceeds the critical value (1.96), we reject the null hypothesis.

Thus, the valid conclusion is:

**One can reject the null hypothesis.**

A simple random sample of size n is drawn from a normally distributed population, and the mean of the sample is x Overbar, while the standard deviation is s. What is the 99% confidence interval for the population mean? Use the table below to help you answer the question.

Confidence Level
90%
95%
99%
z*-score
1.645
1.96
2.58
x Overbar plus-or-minus StartFraction 0.90 times s Over StartRoot n EndRoot EndFraction
x Overbar plus-or-minus StartFraction 0.99 times s Over StartRoot n EndRoot EndFraction
x Overbar plus-or-minus StartFraction 1.645 times s Over StartRoot n EndRoot EndFraction
x Overbar plus-or-minus StartFraction 2.58 times s Over StartRoot n EndRoot EndFraction

To construct a confidence interval for the population mean based on a sample drawn from a normally distributed population, we use the following formula:

\[
\text{Confidence Interval} = \overline{x} \pm z^* \times \frac{s}{\sqrt{n}}
\]

Where:
- \(\overline{x}\) is the sample mean,
- \(z^*\) is the z-score corresponding to the desired confidence level,
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.

### Given Information:
For a **99% confidence level**, the corresponding z-score is provided in your table:
- \( z^* = 2.58 \)

### Confidence Interval Formula for 99% Confidence Level:
Based on the information provided, the confidence interval for the population mean at the 99% confidence level can be expressed as:

\[
\overline{x} \pm \left(2.58 \times \frac{s}{\sqrt{n}}\right)
\]

### Conclusion:
The correct expression for the 99% confidence interval for the population mean from the choices given is:

\[
\overline{x} \pm \frac{s}{\sqrt{n}} \times 2.58
\]

Thus, the correct option from your list is:

**\(\overline{x} \pm \frac{2.58 \times s}{\sqrt{n}}\)**.

In Juneau, Alaska, the 30-year annual snowfall average is 86.7 inches with a standard deviation of 40.4 inches. The last four years saw an average annual snowfall of 115.7 inches, 62.9 inches, 168.5 inches, and 135.7 inches. Hia performs a hypothesis test on this data to determine if the next 30-year norm will have a different average if the trend from the last four years continues. She uses a significance level of 5%. Which of the following is a conclusion that she may make?
The z-statistic is 1.44, so the null hypothesis cannot be rejected.
The z-statistic is 1.68, so the null hypothesis cannot be rejected.
The z-statistic is 1.85, so the null hypothesis should be rejected.
The z-statistic is 4.6, so the null hypothesis should be rejected.

To determine whether Hia can reject the null hypothesis based on the provided information, we first need to perform a hypothesis test. Here’s how we can do that step-by-step:

### Step 1: Formulate the Hypotheses

1. **Null Hypothesis (\(H_0\))**: The average annual snowfall has not changed from the long-term average of 86.7 inches.  
   \( H_0: \mu = 86.7 \)

2. **Alternative Hypothesis (\(H_a\)**: The average annual snowfall has changed from 86.7 inches.  
   \( H_a: \mu \neq 86.7 \)

### Step 2: Calculate the Sample Mean and Standard Deviation

The last four years’ snowfall amounts are:
- 115.7 inches
- 62.9 inches
- 168.5 inches
- 135.7 inches

#### Calculate the sample mean (\(\overline{x}\)):
\[
\overline{x} = \frac{115.7 + 62.9 + 168.5 + 135.7}{4} = \frac{482.8}{4} = 120.7
\]

#### Sample size (\(n\)):
- \(n = 4\)

#### Population standard deviation (\(s\)):
- The population standard deviation is 40.4 inches (since the 30-year average is being used).

### Step 3: Calculate the Z-Statistic

The z-statistic can be calculated using the formula:

\[
z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
\]

Where:
- \(\overline{x} = 120.7\)
- \(\mu = 86.7\)
- \(\sigma = 40.4\)
- \(n = 4\)

#### Calculate the standard error (SE):

\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{40.4}{\sqrt{4}} = \frac{40.4}{2} = 20.2
\]

#### Calculate the z-value:

\[
z = \frac{120.7 - 86.7}{20.2} = \frac{34}{20.2} \approx 1.6842
\]

### Step 4: Compare the Z-Statistic with Critical Values

For a significance level of 5% in a two-tailed test, the critical z-values are approximately ±1.96. 

### Conclusion

- The calculated z-statistic \( \approx 1.68\) does not exceed the critical value of 1.96; therefore, we **do not reject the null hypothesis**.

From the options provided:
1. **The z-statistic is 1.44, so the null hypothesis cannot be rejected.** - Incorrect.
2. **The z-statistic is 1.68, so the null hypothesis cannot be rejected.** - **Correct**.
3. **The z-statistic is 1.85, so the null hypothesis should be rejected.** - Incorrect.
4. **The z-statistic is 4.6, so the null hypothesis should be rejected.** - Incorrect.

Thus, the correct conclusion is:

**The z-statistic is 1.68, so the null hypothesis cannot be rejected.**

The graph below shows three different normal distributions.
3 normal distributions are shown. All 3 are at different points on the axis. They all have different heights and widths.

Which statement must be true?
Each distribution has the same mean and the same standard deviation.
Each distribution has the same mean and a different standard deviation.
Each distribution has a different mean and the same standard deviation.
Each distribution has a different mean and a different standard deviation.

To analyze the statements regarding the three normal distributions displayed in the graph, we can look at the characteristics typically associated with normal distributions:

1. **Mean (\(\mu\))**: This is the point at which the distribution is centered. If the distributions are located at different positions along the horizontal axis, this suggests that they have different means.

2. **Standard Deviation (\(\sigma\))**: This indicates the spread or width of the distribution. If the heights and widths of the distributions are different, it suggests that the standard deviations are also different.

### Analysis of Statements:

- **Each distribution has the same mean and the same standard deviation.**
  - This statement cannot be true because the distributions are at different points on the axis, indicating different means.

- **Each distribution has the same mean and a different standard deviation.**
  - This statement cannot be true for the same reason as above; if they had the same mean, they would all be centered in the same place.

- **Each distribution has a different mean and the same standard deviation.**
  - This statement is also unlikely to be true, as different means with the same standard deviation would typically result in distributions that are similar in shape but appear at different points along the horizontal axis—all having the same spread.

- **Each distribution has a different mean and a different standard deviation.**
  - This statement must be true because:
    - The distributions are located at different positions along the horizontal axis, indicating **different means**.
    - The differences in height and width of the distributions imply **different standard deviations**.

### Conclusion:

The correct statement is:

**Each distribution has a different mean and a different standard deviation.**