Math  /  Data & Statistics

Questionfor a manturing company wishes to reduce the mean time needed serer visit to a customer location in his region. Nationally, this company sets a standard that a customer service visit should last at most four hours, on average. He is concerned that the service technicians in his region may be violating that standard, so he samples 41 customer service visits. In this sample, the mean time needed for a service visit is 4.2 hours with standard deviation 1.6 hours. In a previous example, we determined that the tt-test for the mean should be conducted. a. Set up the null and alternative hypothesis to test the regional manager's claim that the standard set by the company is being violated. b. Calculate the value of the test statistic. c. Find the p-value. d. At the α=0.05\alpha=0.05 level of significance, make a decision whether to reject H0\mathrm{H}_{0}. e. Interpret this conclusion.

Studdy Solution

STEP 1

1. The sample size is n=41 n = 41 .
2. The sample mean is xˉ=4.2 \bar{x} = 4.2 hours.
3. The sample standard deviation is s=1.6 s = 1.6 hours.
4. The national standard for the mean time is μ0=4 \mu_0 = 4 hours.
5. The significance level is α=0.05 \alpha = 0.05 .

STEP 2

1. Set up the null and alternative hypotheses.
2. Calculate the test statistic.
3. Find the p-value.
4. Make a decision based on the p-value and significance level.
5. Interpret the conclusion.

STEP 3

Set up the null and alternative hypotheses:
- Null hypothesis (H0 H_0 ): The mean time is equal to the national standard, μ=4 \mu = 4 . - Alternative hypothesis (Ha H_a ): The mean time is greater than the national standard, μ>4 \mu > 4 .

STEP 4

Calculate the test statistic using the formula for the one-sample t t -test:
t=xˉμ0s/nt = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
Substitute the values:
t=4.241.6/41=0.21.6/41t = \frac{4.2 - 4}{1.6 / \sqrt{41}} = \frac{0.2}{1.6 / \sqrt{41}}
Calculate the denominator:
1.6/410.2491.6 / \sqrt{41} \approx 0.249
Calculate the test statistic:
t0.20.2490.803t \approx \frac{0.2}{0.249} \approx 0.803

STEP 5

Find the p-value for the calculated t t -statistic with n1=40 n-1 = 40 degrees of freedom. This requires consulting a t t -distribution table or using statistical software.
For t0.803 t \approx 0.803 and df=40 df = 40 , the p-value is greater than 0.05.

STEP 6

Make a decision based on the p-value and significance level:
Since the p-value is greater than α=0.05 \alpha = 0.05 , we fail to reject the null hypothesis H0 H_0 .

STEP 7

Interpret the conclusion:
There is not enough statistical evidence to support the claim that the mean service visit time in the region exceeds the national standard of 4 hours.

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