A large television repair service claims that its average repair charge is $124. Suspecting that the average charge is higher, a consumer group randomly choose a sample of 35 statements for tv repairs done by the repair service and found the mean was $125.50 and s = 2.25. Test the claim at the .05 level of significance
Use a one-sample z-test.
Hypotheses:
Ho: µ = 124 -->null hypothesis
Ha: µ > 124 -->alternate hypothesis
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
With your data:
z = (125.50 - 124)/(2.25/√35) = ?
Finish the calculation.
Check a z-table at .05 level of significance for a one-tailed test.
If the z-test statistic exceeds the critical value from the z-table, reject the null. If the z-test statistic does not exceed the critical value from the z-table, do not reject the null.
I hope this will help get you started.
To test the claim at the 0.05 level of significance, we can use a one-sample t-test.
Step 1: State the hypothesis
The null hypothesis (H0) is that the average repair charge is $124, while the alternative hypothesis (Ha) is that the average repair charge is higher than $124.
H0: μ = $124
Ha: μ > $124
Step 2: Determine the significance level and find the critical value
The significance level (α) is given as 0.05. Since this is a one-tailed test (upper-tailed), we need to find the critical value associated with the given significance level.
For a one-tailed test, with a significance level of 0.05, and degrees of freedom (df) = n - 1 = 35 - 1 = 34, the critical value can be found using t-table or a t-distribution calculator.
The critical value at 0.05 level of significance and 34 degrees of freedom is approximately 1.6909.
Step 3: Calculate the test statistic
The test statistic is calculated using the formula:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the population mean (claimed average repair charge), s is the sample standard deviation, and n is the sample size.
In this case, x̄ = $125.50, μ = $124, s = 2.25, and n = 35.
Plugging in the values, we get:
t = (125.50 - 124) / (2.25 / √35)
Step 4: Make a decision and draw a conclusion
Compare the calculated test statistic to the critical value.
If the calculated test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Let's calculate the test statistic:
t = (125.50 - 124) / (2.25 / √35)
= 1.50 / (2.25 / √35)
= 1.50 / (2.25 / 5.916)
= 1.50 / 0.3807
= 3.9400
Since 3.9400 is greater than the critical value of 1.6909, we reject the null hypothesis.
Conclusion: Based on the given sample data, there is enough evidence to suggest that the average repair charge by the television repair service is higher than $124 at the 0.05 level of significance.