Solved on Dec 03, 2023

Test whether the mean retirement age of women executives is different from the reported 62.7 years, given a sample of 80 with $\sigma = 4.7$ years.
$\begin{array}{l} H_{0}: \mu = 62.7 \\ H_{a}: \mu \neq 62.7 \end{array}$

STEP 1

Assumptions
1. The reported mean age of retirement for women executives is 62.7 years.
2. The sample mean age of retirement for 80 recently retired women executives is 63.8 years.
3. The population standard deviation is 4.7 years.
4. The level of significance for the test is 0.05.

STEP 2

State the null and alternative hypotheses for the test.
The null hypothesis (H0) is that the mean age of retirement for women executives is 62.7 years. The alternative hypothesis (Ha) is that the mean age of retirement for women executives is not 62.7 years.
$\begin{array}{l} H_{0}: \mu=62.7 \\ H_{a}: \mu \neq 62.7 \end{array}$

STEP 3

Calculate the standard error of the mean. The standard error of the mean is the standard deviation divided by the square root of the sample size.
$SE = \frac{\sigma}{\sqrt{n}}$

STEP 4

Plug in the values for the standard deviation and the sample size to calculate the standard error.
$SE = \frac{4.7}{\sqrt{80}}$

STEP 5

Calculate the standard error.
$SE = \frac{4.7}{\sqrt{80}} = 0.526$

STEP 6

Calculate the test statistic (Z-score). The Z-score is the difference between the sample mean and the population mean divided by the standard error.
$Z = \frac{\bar{x} - \mu}{SE}$

STEP 7

Plug in the values for the sample mean, the population mean, and the standard error to calculate the Z-score.
$Z = \frac{63.8 - 62.7}{0.526}$

STEP 8

Calculate the Z-score.
$Z = \frac{63.8 - 62.7}{0.526} = 2.09$

STEP 9

Compare the calculated Z-score with the critical Z-score for the given level of significance (0.05). The critical Z-score for a two-tailed test at the 0.05 level of significance is approximately ±1.96.
Since the calculated Z-score (2.09) is greater than the critical Z-score (1.96), we reject the null hypothesis. There is sufficient evidence to support the organization's belief that the mean age of retirement for women executives is not 62.7 years.

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Test whether the mean retirement age of women executives is different from the reported 62.7 years, given a sample of 80 with $\sigma = 4.7$ years.
$\begin{array}{l} H_{0}: \mu = 62.7 \\ H_{a}: \mu \neq 62.7 \end{array}$

STEP 1

Assumptions
1. The reported mean age of retirement for women executives is 62.7 years.
2. The sample mean age of retirement for 80 recently retired women executives is 63.8 years.
3. The population standard deviation is 4.7 years.
4. The level of significance for the test is 0.05.

STEP 2

STEP 3

Calculate the standard error of the mean. The standard error of the mean is the standard deviation divided by the square root of the sample size.
$SE = \frac{\sigma}{\sqrt{n}}$

STEP 4

Plug in the values for the standard deviation and the sample size to calculate the standard error.
$SE = \frac{4.7}{\sqrt{80}}$

STEP 5

Calculate the standard error.
$SE = \frac{4.7}{\sqrt{80}} = 0.526$

STEP 6

Calculate the test statistic (Z-score). The Z-score is the difference between the sample mean and the population mean divided by the standard error.
$Z = \frac{\bar{x} - \mu}{SE}$

STEP 7

Plug in the values for the sample mean, the population mean, and the standard error to calculate the Z-score.
$Z = \frac{63.8 - 62.7}{0.526}$

STEP 8

Calculate the Z-score.
$Z = \frac{63.8 - 62.7}{0.526} = 2.09$

STEP 9

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Test whether the mean retirement age of women executives is different from the reported 62.7 years, given a sample of 80 with σ=4.7\sigma = 4.7σ=4.7 years.H0:μ=62.7Ha:μ≠62.7 \begin{array}{l} H_{0}: \mu = 62.7 \\ H_{a}: \mu \neq 62.7 \end{array} H0​:μ=62.7Ha​:μ=62.7​

STEP 1

Assumptions1. The reported mean age of retirement for women executives is 62.7 years.2. The sample mean age of retirement for 80 recently retired women executives is 63.8 years.3. The population standard deviation is 4.7 years.4. The level of significance for the test is 0.05.

STEP 2

STEP 3

Calculate the standard error of the mean. The standard error of the mean is the standard deviation divided by the square root of the sample size.SE=σnSE = \frac{\sigma}{\sqrt{n}}SE=n​σ​

STEP 4

Plug in the values for the standard deviation and the sample size to calculate the standard error.SE=4.780SE = \frac{4.7}{\sqrt{80}}SE=80​4.7​

STEP 5

Calculate the standard error.SE=4.780=0.526SE = \frac{4.7}{\sqrt{80}} = 0.526SE=80​4.7​=0.526

STEP 6

Calculate the test statistic (Z-score). The Z-score is the difference between the sample mean and the population mean divided by the standard error.Z=xˉ−μSEZ = \frac{\bar{x} - \mu}{SE}Z=SExˉ−μ​

STEP 7

Plug in the values for the sample mean, the population mean, and the standard error to calculate the Z-score.Z=63.8−62.70.526Z = \frac{63.8 - 62.7}{0.526}Z=0.52663.8−62.7​

STEP 8

Calculate the Z-score.Z=63.8−62.70.526=2.09Z = \frac{63.8 - 62.7}{0.526} = 2.09Z=0.52663.8−62.7​=2.09

STEP 9

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Test whether the mean retirement age of women executives is different from the reported 62.7 years, given a sample of 80 with $\sigma = 4.7$ years.
$\begin{array}{l} H_{0}: \mu = 62.7 \\ H_{a}: \mu \neq 62.7 \end{array}$

Assumptions
1. The reported mean age of retirement for women executives is 62.7 years.
2. The sample mean age of retirement for 80 recently retired women executives is 63.8 years.
3. The population standard deviation is 4.7 years.
4. The level of significance for the test is 0.05.

Calculate the standard error of the mean. The standard error of the mean is the standard deviation divided by the square root of the sample size.
$SE = \frac{\sigma}{\sqrt{n}}$

Plug in the values for the standard deviation and the sample size to calculate the standard error.
$SE = \frac{4.7}{\sqrt{80}}$

Calculate the standard error.
$SE = \frac{4.7}{\sqrt{80}} = 0.526$

Calculate the test statistic (Z-score). The Z-score is the difference between the sample mean and the population mean divided by the standard error.
$Z = \frac{\bar{x} - \mu}{SE}$

Plug in the values for the sample mean, the population mean, and the standard error to calculate the Z-score.
$Z = \frac{63.8 - 62.7}{0.526}$

Calculate the Z-score.
$Z = \frac{63.8 - 62.7}{0.526} = 2.09$