Questions: One cable company claims that it has excellent customer service. In fact, the company advertises that a technician will arrive within 50 minutes after a service call is placed. One frustrated customer believes this is not accurate, claiming that it takes over 50 minutes for the cable technician to arrive. The customer asks a simple random sample of 16 other cable customers how long it has taken for the cable technician to arrive when they have called for one. The sample mean for this group is 55.1 minutes with a standard deviation of 10.6 minutes. Assume that the population distribution is approximately normal. Test the customer's claim at the 0.005 level of significance. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ=50 Ha: μ ≠ 50

One cable company claims that it has excellent customer service. In fact, the company advertises that a technician will arrive within 50 minutes after a service call is placed. One frustrated customer believes this is not accurate, claiming that it takes over 50 minutes for the cable technician to arrive. The customer asks a simple random sample of 16 other cable customers how long it has taken for the cable technician to arrive when they have called for one. The sample mean for this group is 55.1 minutes with a standard deviation of 10.6 minutes. Assume that the population distribution is approximately normal. Test the customer's claim at the 0.005 level of significance.
Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below.
H0: μ=50
Ha: μ ≠ 50

Transcript text: One cable company claims that it has excellent customer service. In fact, the company advertises that a technician will arrive within 50 minutes after a service call is placed. One frustrated customer believes this is not accurate, claiming that it takes over 50 minutes for the cable technician to arrive. The customer asks a simple random sample of 16 other cable customers how long it has taken for the cable technician to arrive when they have called for one. The sample mean for this group is 55.1 minutes with a standard deviation of 10.6 minutes. Assume that the population distribution is approximately normal. Test the customer's claim at the 0.005 level of significance. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. \[ \begin{array}{l} H_{0}: \mu=50 \\ H_{a}: \mu ـ 50 \end{array} \]

Solution

Solution Steps

Step 1: State the Hypotheses

We are testing the following hypotheses:

Null Hypothesis \( H_0: \mu = 50 \)
Alternative Hypothesis \( H_a: \mu > 50 \)

Step 2: Calculate the Standard Error

The standard error \( SE \) is calculated as follows: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{10.6}{\sqrt{16}} = 2.65 \]

Step 3: Calculate the Test Statistic

The test statistic \( t_{\text{test}} \) is calculated using the formula: \[ t_{\text{test}} = \frac{\bar{x} - \mu_0}{SE} = \frac{55.1 - 50}{2.65} = 1.9245 \]

Step 4: Calculate the P-value

For a right-tailed test, the P-value is calculated as: \[ P = 1 - T(z) = 0.0367 \]

Step 5: Conclusion

We compare the P-value with the significance level \( \alpha = 0.005 \):

Since \( 0.0367 > 0.005 \), we fail to reject the null hypothesis.

Final Answer

The conclusion is that there is not enough evidence to support the claim that the mean time for the cable technician to arrive is greater than 50 minutes.

\(\boxed{\text{Fail to reject } H_0}\)

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