An auto maker estimates that the mean gas mileage of its sport utility vehicle is 20 miles per gallon. A random sample of 8 such vehicles had a mean of 18 per gallon and a standard deviation of 5 miles per gallon. At á=0.05, can you reject the auto maker’s claim that the mean gas mileage of its sports
utility vehicle is 20 miles per gallon? Assume the population is normally distributed.
You can use a t-test for this problem since the sample size is small.
Using your data in the formula:
t = (18 - 20)/(5/√8) = ?
Finish the calculation.
Using a t-table at 0.05 level of significance for a two-tailed test (the test is two-tailed because the alternative hypothesis would not specify a direction) at 7 degrees of freedom (df = n - 1 = 8 - 1 = 7), find your critical or cutoff value to reject the null. If the t-test statistic calculated above does not exceed the critical value you found in the t-table, you cannot reject the null. If the t-test statistic exceeds the critical value from the t-table, reject the null.
I hope this will help get you started.
of 500 employee ,200 participating i n companies profit sharing plan (p),250 having major medical insurance coverage (m) and 50 find the probibility +will not be participant in either program
of 500 employee ,200 participating i n companies profit sharing plan (p),250 having major medical insurance coverage (m) and 50 participated in both program
find the probibility will not be participant in either program
To determine whether you can reject the auto maker's claim that the mean gas mileage of its sports utility vehicle is 20 miles per gallon, you can conduct a hypothesis test. The hypothesis test will help you determine if the provided sample mean of 18 miles per gallon is statistically significantly different from the claimed mean of 20 miles per gallon.
Here's how you can conduct the hypothesis test:
Step 1: State the hypotheses:
The null hypothesis, denoted as H0, represents the assumption that the mean gas mileage is 20 miles per gallon. The alternative hypothesis, denoted as Ha, represents the claim that the mean gas mileage is not 20 miles per gallon.
H0: μ = 20
Ha: μ ≠ 20
Step 2: Set the significance level (α):
The significance level represents the probability of rejecting the null hypothesis, assuming it is true. In this case, the significance level is stated as α = 0.05.
Step 3: Compute the test statistic:
The test statistic for comparing means is the t-statistic. Since the population standard deviation is not known, you need to use the t-test statistic. The formula for the t-statistic is:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean (18), μ is the claimed mean (20), s is the sample standard deviation (5), and n is the sample size (8).
Calculating the t-statistic:
t = (18 - 20) / (5 / √8)
t = -2 / (5 / √8)
t ≈ -1.7889
Step 4: Determine the critical value(s):
Since the alternative hypothesis is two-tailed (μ ≠ 20), you need to find the critical values for a two-tailed test at α = 0.05. The critical values can be found using a t-distribution table or a statistical calculator. In this case, with a sample size of 8 and a significance level of 0.05, the critical values are approximately ±2.306.
Step 5: Make a decision:
Compare the test statistic (t) to the critical value(s). If the absolute value of the test statistic is greater than the critical value, you reject the null hypothesis. Otherwise, you fail to reject the null hypothesis.
In this case, |t| ≈ 1.7889 < 2.306, so the test statistic does not exceed the critical value. Therefore, you fail to reject the null hypothesis.
Step 6: State the conclusion:
Based on the results, at α = 0.05, there is not enough evidence to reject the auto maker's claim that the mean gas mileage of its sports utility vehicle is 20 miles per gallon.