For question A, the drop down options are : (1) Greater than 1,505 (2) Greater t
ID: 3353171 • Letter: F
Question
For question A, the drop down options are : (1) Greater than 1,505 (2) Greater than or equal to 1,505 (3) equal to 1,505 (4) less than or equal to 1,505 (5) less than 1,505 (6) not equal to 1,505
For question C the drop down options for p-value are ;
(1) less that .005 (2) between .005 and .01 (3)between .01 and .025 (4) between.025 and .05 (5) between .05 and .10 (6) between .10 and .20 (7) greater than .20
(Exercise 9.33 (Algorithmic) The mean annual premium for automobile insurance in the United States is $1,505. Being from Pennsylvania, you believe automobile insurance is cheaper there and wish to develop statistical support for your opinion. A sample of 25 automobile insurance policies from the state of Pennsylvania showed a mean annual premium of $1,446 with a standard deviation of s $155. If required, enter negative values as negative numbers. a. Develop a hypothesis test that can be used to determine whether the mean annual premium in Pennsylvania is lower than the national mean annual premium. Ho: 1 equal to 1,505 Ha: less than 1,505 b. What is a point estimate of the difference between the mean annual premium in Pennsylvania and the national mean (to the nearest dollar)? C. At o = .05, test for a significant difference by completing the following. Calculate the value of the test statistic (to 2 decimals). 1.90 The p-value is between.05 and.10 What is your conclusion? The population mean automobile premium is lower in Pennsylvania than the national mean.Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: > 1505
Alternative hypothesis: < 1505
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 31
DF = n - 1
D.F = 24
t = (x - ) / SE
t = - 1.90
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of - 1.90.
Thus the P-value in this analysis is 0.035.
Interpret results. Since the P-value (0.035) is less than the significance level (0.05), we have to reject the null hypothesis.
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