As the administrator of a regional school system, you are concerned with insurin
ID: 3317555 • Letter: A
Question
As the administrator of a regional school system, you are concerned with insuring that your students are receiving everything that they need in order to be successful in school. In particular, you recognize the value of good nutrition as it relates to academic performance. Therefore, you would like to apply for a grant to help fund the “free and reduced lunch” program. The grant provides additional funding to school systems that average more than 4,250 students per district who are receiving free or reduced lunches. You randomly sample 50 school districts from your system and record the number of students in each district who are receiving free or reduced lunches. The results are summarized below.
At the 10% significance level, do the data suggest that you will be able to win the grant for your school system?
# of students receiving free or reduced lunch
Mean
5094.5
Standard Error
484.95
Median
4496
Standard Deviation
3429.14
Sample Variance
11759026.09
Kurtosis
2.23
Skewness
1.43
Range
15610
Minimum
698
Maximum
16308
Sum
254725
Count
50
Would the p-value for this hypothesis test be greater than .10 or less than .10? Why?
Would your results change if you were required to conduct the hypothesis test at the 5% significance level? Why or why not?
# of students receiving free or reduced lunch
Mean
5094.5
Standard Error
484.95
Median
4496
Standard Deviation
3429.14
Sample Variance
11759026.09
Kurtosis
2.23
Skewness
1.43
Range
15610
Minimum
698
Maximum
16308
Sum
254725
Count
50
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 4250
Alternative hypothesis: > 4250
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 0.10. 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 = 484.95
DF = n - 1
D.F = 49
t = (x - ) / SE
t = 1.74
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 1.74.
Thus the P-value in this analysis is 0.041.
Interpret results. Since the P-value (0.041) is less than the significance level (0.10), we have to reject the null hypothesis.
At 5% significance level.
Thus the P-value in this analysis is 0.041.
Interpret results. Since the P-value (0.041) is less than the significance level (0.05), we have to reject the null hypothesis.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.