Question

A researcher has administered an optimism test for graduating college students for the past several years. In the current yearthe researcher obtains a sample of n = 36 people and asks each person to complete the optimism test (higher scores indicate higher levels of optimism about the future). The scores from the sample produce a mean of M=24wit4ss=2835 . Over the past few years the optimism test is known to have a mean of w = 20 Do the sample data support the conclusion that the current graduating students have a different level optimism than in previous years?

Answer 1

Given : Sample size=n=36

Sample mean =$\bar{X}=M=24$

Sample standard deviation=s=$\sqrt{SS/n-1}=\sqrt{2835/35}=9$

Hypothesized value=$\mu_0=20$

Assume that , Significance level=

= 0,05

Hypothesis: $H_0:\mu=20$ Vs $H_a:\mu\neq 20$

The test is two-tailed test.

Since , the population standard deviation is unknown.

Therfore , use t-distribution.

df=degrees of freedom=n-1=36-1=35

The test statistic is ,

$t_{stat}=\frac{\bar{X}-\mu_0}{s/\sqrt{n}}=\frac{24-20}{9/\sqrt{36}}=2.6667$

The critical values are ,

$t_{df,\alpha/2}=t_{35,0.05/2}=2.0301$ ; The Excel function is , =TINV(0.05,35)

" Do not re Ho to Rejection region Rejection region

Decision : Here , the value of the test statistic lies in the rejection region.

Therefore , reject Ho.

Conclusion : Hence ,  the sample data support the conclusion that the current graduating students have a different level optimism than in previous years.