The average monthly Social Security benefit for widows and widowers in Texas was

Question

The average monthly Social Security benefit for widows and widowers in Texas was $940 in 2005. A social welfare advocate from a rural Texas county believed that the 2005 average monthly benefit for widows and widowers in her county was significantly less than the overall Texas average. To test this assumption, the advocate randomly sampled 100 widow / widower monthly benefits and found the sample mean to be $915 with a standard deviation of $90.

10. What is the form of the alternative hypothesis needed to test the social welfare advocates belief? a. Ha: µ < 915 b. Ha: p < 940 c. Ha: µ < 940

11. What is the value of the test statistic? a. -2.778 b. -1.96 c. -915

12. Under the null hypothesis, will the test statistic have a normal (z) distribution or a tdistribution? a. Normal (z) distribution b. t-distribution with n-1 degrees of freedom c. Not enough information to determine the distribution of the test statistic.

13. The p-value for this test was reported to be 0.007. What conclusion do you reach? Assume a significance level of 0.01. a. Reject Ho and conclude that the data indicate that the widow / widower monthly benefits in the rural county are less than the average benefit for the state of Texas. b. Fail to reject Ho and conclude that the data do not provide sufficient evidence to conclude that the widow / widower benefit for the rural county is less than the average benefit for the state of Texas

Explanation / Answer

10) the alternate hypothesis will be Ha<940 because the mean is given to be 940 and in the sample we got mean = 915

11) here we need to find the value of test static

z = (x-u)/(s/sqrt(n))

= (915-940)/(90/sqrt(100) = -2778

12) it will b a normal distribution as the number of the sample is greater than 30.

13) p value is 0.007 which is in significant with respect to 0.01 therefore the null hypothesis will be rejected and the option A is true.

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