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CareerBuilder conducted a survey of 2245 employers asking if they had fired an employee for shopping online while at work. 257 employers said they had fired an employee for online shopping while at work. A manager at a local recruiting agency claims that more than 10% of employers fire an employee for online shopping while at work. Does the sample data provide evidence to support this manager’s claim at α=.05?


1. A. Is this a test for a population proportion or a population mean?
B. What distribution is used to conduct this test?
C. Is this a left-tailed, right-tailed, or two-tailed test?

2. State AND verify all assumptions required for this test.
[HINT: This test should have four assumptions to be verified.]

5 answers

To address the questions regarding the manager's claim and the survey results, let's analyze the data step by step.

1A. Is this a test for a population proportion or a population mean?

This is a test for a population proportion. The manager's claim is concerning the proportion of employers who fire employees for online shopping while at work, which is a categorical outcome (fired or not fired).

1B. What distribution is used to conduct this test?

The distribution used to conduct this test is the normal distribution, specifically the sampling distribution of the sample proportion, which can be approximated by the normal distribution due to the sample size being large.

1C. Is this a left-tailed, right-tailed, or two-tailed test?

This is a right-tailed test. The manager claims that more than 10% of employers fire employees for online shopping, which corresponds to testing whether the proportion is greater than 0.10.

2. State AND verify all assumptions required for this test.

To conduct a hypothesis test for a population proportion, we need to verify the following assumptions:

  1. Random Sampling: The sample should be a simple random sample from the population.

    • Verification: Verify that the 2245 employers surveyed were chosen randomly from the population of all employers.

  2. Independence: The responses from each individual in the sample should be independent of each other.

    • Verification: The survey should not have any design that would bias or influence the answers of employers (e.g., employers working in a similar industry may share common behaviors affecting independence).

  3. Sample Size: The sample size should be large enough to invoke the Central Limit Theorem. Specifically, both \( np \) and \( n(1-p) \) should be at least 5 for the normal approximation to be valid, where \( n \) is the sample size and \( p \) is the sample proportion.

    • Verification:
      • Calculate the sample proportion \( \hat{p} = \frac{257}{2245} \approx 0.1145 \).
      • Check \( np \) and \( n(1-p) \):
        • \( np = 2245 \times 0.1145 \approx 257 \)
        • \( n(1-p) = 2245 \times (1 - 0.1145) \approx 1988 \)
      • Both values are greater than 5, thus this assumption is satisfied.

  4. Proportion of Interest: The assumed population proportion under the null hypothesis should be a realistic value.

    • Verification: The null hypothesis is \( H_0: p = 0.10 \). This is a stated proportion and thus is a reasonable value to test against.

If all these assumptions are met, we can proceed with the hypothesis test for the population proportion to evaluate the manager's claim.

State the null and alternate hypotheses for this test: (use correct symbols and format!)

Null hypothesis : Ho:
Alternate hypothesis : Ha:
Run the correct hypothesis test and provide the information below. Give the correct symbols AND numeric value of each of the following (round answers to 3 decimal places).

Test Statistic:
Critical value [HINT: this is NOT α] :
p-value :

To test the manager's claim about the proportion of employers who fire employees for online shopping, we will set up the null and alternate hypotheses and then calculate the test statistic, critical value, and p-value.

Null and Alternate Hypotheses

  • Null Hypothesis (\(H_0\)): \(p = 0.10\) (The proportion of employers who fire an employee for online shopping is 10%.)
  • Alternate Hypothesis (\(H_a\)): \(p > 0.10\) (The proportion of employers who fire an employee for online shopping is greater than 10%.)

Information Required to Conduct the Test

  1. Sample Size (\(n\)): 2245
  2. Number of successes (fired employees): 257
  3. Sample Proportion (\(\hat{p}\)): \[ \hat{p} = \frac{257}{2245} \approx 0.1145 \]
  4. Null Hypothesis Proportion (\(p_0\)): 0.10

Test Statistic

The test statistic for a proportion can be calculated using the formula: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] Calculating the standard deviation: \[ \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.10(1 - 0.10)}{2245}} = \sqrt{\frac{0.10 \times 0.90}{2245}} \approx \sqrt{\frac{0.09}{2245}} \approx 0.006294 \]

Now substituting values into the z formula: \[ z = \frac{0.1145 - 0.10}{0.006294} \approx \frac{0.0145}{0.006294} \approx 2.307 \]

Critical Value

For a right-tailed test at \(\alpha = 0.05\), we find the critical value from the z-table: \[ z_{\text{critical}} \approx 1.645 \]

p-value

To find the p-value, we can use the z-score we calculated: Using a standard normal distribution table or calculator to find the area to the right of \(z = 2.307\): \[ p\text{-value} \approx P(Z > 2.307) \approx 0.0105 \]

Summary Results

  • Test Statistic: \[ z \approx 2.307 \]
  • Critical Value: \[ z_{\text{critical}} \approx 1.645 \]
  • p-value: \[ \text{p-value} \approx 0.0105 \]

Rounded Answers

  • Test Statistic: \(2.307\)
  • Critical Value: \(1.645\)
  • p-value: \(0.011\) (rounded to three decimal places)

In summary, \(z \approx 2.307\), the critical value is \(1.645\), and the p-value is \(0.011\). Since the p-value is less than \(\alpha = 0.05\), we reject the null hypothesis and conclude that there is significant evidence to support the manager's claim that more than 10% of employers fire an employee for online shopping while at work.

6. Find a 95% confidence interval for the proportion of employers who fire employees shopping online at work. Interpret this interval in context to the manager’s claim above. (Hint: Does 10% lie in the interval?) Show calculator command and/or formula used to get answer.

To find a 95% confidence interval for the proportion of employers who fire employees for shopping online at work, we use the formula for a confidence interval for a population proportion:

\[ \text{CI} = \hat{p} \pm z_{\frac{\alpha}{2}} \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]

Where:

  • \(\hat{p}\) = sample proportion = \(\frac{257}{2245} \approx 0.1145\)
  • \(n\) = sample size = 2245
  • \(z_{\frac{\alpha}{2}}\) for a 95% confidence level ≈ 1.96 (the critical value corresponding to \(\alpha = 0.05\))

Step-by-step Calculation

  1. Calculate the standard error (SE): \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.1145(1 - 0.1145)}{2245}} \] \[ SE = \sqrt{\frac{0.1145 \times 0.8855}{2245}} \approx \sqrt{\frac{0.10170775}{2245}} \approx \sqrt{0.00004524} \approx 0.00671 \]

  2. Calculate the margin of error (ME): \[ ME = z_{\frac{\alpha}{2}} \times SE = 1.96 \times 0.00671 \approx 0.01315 \]

  3. Calculate the confidence interval: \[ \text{CI} = \hat{p} \pm ME = 0.1145 \pm 0.01315 \] \[ \text{Lower Limit} = 0.1145 - 0.01315 \approx 0.10135 \] \[ \text{Upper Limit} = 0.1145 + 0.01315 \approx 0.12765 \]

Final Confidence Interval

Thus, the 95% confidence interval for the proportion of employers who fire employees for online shopping while at work is: \[ (0.101, 0.128) \]

Interpretation of the Interval

This means we are 95% confident that the true proportion of employers who fire an employee for shopping online while at work is between approximately 10.1% and 12.8%.

Contextual Interpretation in Relation to the Manager’s Claim

  • The manager claimed that more than 10% of employers fire employees for online shopping at work.
  • The lower limit of the confidence interval is approximately 10.1%, which is just slightly above 10%. Since this interval does not include values below 10% and starts just above it, it supports the manager's claim that more than 10% of employers do fire employees for shopping online while at work. This provides statistical validity to the manager's assertion that the proportion is likely greater than 10%.