The director of admissions at the UMUC, is concerned about the high cost of text
ID: 3175486 • Letter: T
Question
The director of admissions at the UMUC, is concerned about the high cost of textbooks for the students each semester. A sample of 25 students enrolled in the university indicates that x(bar)=$315.40 and $43.20.
Using a significance level .1, what are the boundaries of the confidence interval?
Using the .10 level of significance, is there evidence that the population mean is above $300?
Using the .05 level of significance, is there evidence that the population mean is above $300?
B)
Explain the difference between parts A and C. In both parts we are asked whether or not there is evidence that the population mean was over $300.
In part A:x(bar)=$315.40; s=$43.2; Sample size = 25
In part C:x(bar)= $315.40; s$75; Sample size = 25
C)
Person
Before
After
1
176
164
2
192
191
3
185
176
4
177
176
5
196
185
6
178
169
7
196
196
8
181
172
9
158
158
10
201
193
11
191
185
12
193
189
13
176
175
14
212
210
15
177
173
16
183
180
17
210
204
18
198
192
19
157
152
20
213
200
21
161
161
22
177
166
23
210
203
24
192
186
25
178
170
(The weight in the column labeled “After” represents their weights six months later and “Before” represents their weight at the start of the six-month period.) The director used .05 as the significance level.
Us Excel to test. For each paired difference, compute After-Before in Data Analysis, t-Test, Paired Two Sample for means, select the After data for Variable 1 Range. Note that the critical value output by Data Analysis for this test is always positive. In this problem, the sign of the critical value is negative corresponding to a 1-tailed test with lower reject region and negative lower critical value.
In a one tail test HO is rejected if P(T<=t) one tail<significance level
What are the Null and Alternate Hypotheses?
What is the t-statistic (t-score) in the Excel output?
What is the t Critical one-tail in the Excel output?
What is your conclusion?
Person
Before
After
1
176
164
2
192
191
3
185
176
4
177
176
5
196
185
6
178
169
7
196
196
8
181
172
9
158
158
10
201
193
11
191
185
12
193
189
13
176
175
14
212
210
15
177
173
16
183
180
17
210
204
18
198
192
19
157
152
20
213
200
21
161
161
22
177
166
23
210
203
24
192
186
25
178
170
Explanation / Answer
Solution:-
A)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 300
Alternative hypothesis: > 300
Note that these hypotheses constitute a one-tailed test.
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 = 8.64
DF = n - 1 = 25 - 1
D.F = 24
t = (x - ) / SE
t = 1.78
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 1.78. We use the t Distribution Calculator to find P(t > 1.78) = 0.0439
Thus the P-value in this analysis is 0.0439
Interpret results. Since the P-value (0.0439) is less than the significance level (0.10), we have to reject the null hypothesis.
Conclusion:- From the above test we have sufficient evidence in the favor of the claim that the population mean is above $300.
At 0.05 level of significance.
Interpret results. Since the P-value (0.0439) is less than the significance level (0.05), we have to reject the null hypothesis.
Conclusion:- From the above test we have sufficient evidence in the favor of the claim that the population mean is above $300.
B)
For part C:x(bar)= $315.40; s = $75; Sample size = 25
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 300
Alternative hypothesis: > 300
Note that these hypotheses constitute a one-tailed test.
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 = 15
DF = n - 1 = 25 - 1
D.F = 24
t = (x - ) / SE
t = 1.027
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 1.027. We use the t Distribution Calculator to find P(t > 1.027) = 0.15733
Thus the P-value in this analysis is 0.15733
Interpret results. Since the P-value (0.15733) is greater than the significance level (0.10), we have to accept the null hypothesis.
Conclusion:- From the above test we do not have sufficient evidence in the favor of the claim that the population mean is above $300.
At 0.05 level of significance.
Interpret results. Since the P-value (0.15733) is greater than the significance level (0.05), we have to accept the null hypothesis.
Conclusion:- From the above test we do not have sufficient evidence in the favor of the claim that the population mean is above $300.