Question

The Department of Education would like to test the hypothesis that the average debt load of graduating students with a bachelor's degree is equal to $17,600. A random sample of 28 students had an average debt load of $18,800. It is believed that the population standard deviation for student debt load is $4800. The a is set to 0.05. The confidence interval for this hypothesis test would be ($17,022.05, $20,577.95] ($14,352.73, $20,847.37] O $16,513.53, $18,686.47] [$15,072.93, $20,127.07] Question 20 1 pts evidence that the null If we reject a null hypothesis at the 1% significance level, then we have hypothesis is false. For this problem create your own t-test and state your null hypothesis, alternative hypothesis Draw out the normal distribution curve and the critical region. You can assume a two tail test. Label the critical region. weak strong O non existent

Answer 1

1. Answer: ($17022.05, 20577.95)

2. Answer: STRONG

NOTE: For second question, random values have been chosen to Calculate critical values and draw normal distribution.

Dete $ ). Average debt CX) = 18,800 Standead deviation (o)= 4800 Given significance level (@=0.05 do confidence Level - 95% Value of 2 at 95% confidence level = 1.96 Sample size (n) = 28 Upper confidence interval →X+ ZXJ +(1292 ► 20,577.95 Lower confidence interval= X-zxc (1.962 Sn >> 18,800 + 1.96% 4800 √28 un -> 18, 800 - 1.96% 4800 > 17, 02205 Confidence internal >> XLRDADOS Confidence interval => (17022:05, 20 572 95) Scanned by TapScanner

Date Page Null hypothesis : Ho: M=50 Alternative hypothesis : Hill 350 Given nce level 0.01 significance Therefore, confidence level 9.99% & Sample Size (n) = 9 Degree of freedom 37 1-1 = 9-1 - 8 * This is a two tailed test. and To find the critical value, look for two tailed t-test table & find value off critical Critical values at degree of 10.01 These fose Critical values = -3.355, +3.355 X = 62 Standard deviation ($) =90 significance level & of Let I n = 9. t-stat= X-u s/sn Scanned by TapScanner

t-stat = 62.50 9119 & lg & 4 Accepted Region -3.355 3.355 t-stat Since value of t-stat doen't lie in the accepted region. have STRONG reject & state that null null hylpothesis Thybothesis false. Therefore, عليا levidence to Scanned by TapScanner