a ) using excel>addin>phstat>confidence interval
we have
Confidence Interval Estimate for the Mean | |
Data | |
Sample Standard Deviation | 5.033222957 |
Sample Mean | 222.6666667 |
Sample Size | 12 |
Confidence Level | 95% |
Intermediate Calculations | |
Standard Error of the Mean | 1.452966315 |
Degrees of Freedom | 11 |
t Value | 2.2010 |
Interval Half Width | 3.1980 |
Confidence Interval | |
Interval Lower Limit | 219.47 |
Interval Upper Limit | 225.86 |
95% confidence interval for the mean mass of bags of Calbie Chips is (219.47,225.86)
2 ) using excel>addin>phstat>two sample test
we have
Pooled-Variance t Test for the Difference Between Two Means | ||||
(assumes equal population variances) | ||||
Data | Confidence Interval Estimate | |||
Hypothesized Difference | 0 | for the Difference Between Two Means | ||
Level of Significance | 0.05 | |||
Population 1 Sample | Data | |||
Sample Size | 12 | Confidence Level | 90% | |
Sample Mean | 78.8 | |||
Sample Standard Deviation | 8.5 | Intermediate Calculations | ||
Population 2 Sample | Degrees of Freedom | 19 | ||
Sample Size | 9 | t Value | 1.7291 | |
Sample Mean | 86 | Interval Half Width | 46.2762 | |
Sample Standard Deviation | 93 | |||
Confidence Interval | ||||
Intermediate Calculations | Interval Lower Limit | -53.4762 | ||
Population 1 Sample Degrees of Freedom | 11 | Interval Upper Limit | 39.0762 | |
Population 2 Sample Degrees of Freedom | 8 | |||
Total Degrees of Freedom | 19 | |||
Pooled Variance | 3683.5132 | |||
Standard Error | 26.7626 | |||
Difference in Sample Means | -7.2000 | |||
t Test Statistic | -0.2690 | |||
Two-Tail Test | ||||
Lower Critical Value | -2.0930 | |||
Upper Critical Value | 2.0930 | |||
p-Value | 0.7908 |
the 90% confidence interval for the difference in midterm test scores of the two sections is (-53.4762,39.0762)
since confidence interval contains 0 so we do not have sufficient evidence to conclude that that one section of students are ”smarter” than the other section.
Ans 3 ) using excel>addin>phstat>two sample test >z test for difference of proportion
we have
Z Test for Differences in Two Proportions | ||||
Data | Confidence Interval Estimate | |||
Hypothesized Difference | 0 | of the Difference Between Two Proportions | ||
Level of Significance | 0.05 | |||
US | Data | |||
Number of Items of Interest | 33288 | Confidence Level | 92% | |
Sample Size | 671493 | |||
UK | Intermediate Calculations | |||
Number of Items of Interest | 14606 | Z Value | -1.7507 | |
Sample Size | 109754 | Std. Error of the Diff. between two Proportions | 0.0011 | |
Interval Half Width | 0.0019 | |||
Intermediate Calculations | ||||
Group 1 Proportion | 0.049573115 | Confidence Interval | ||
Group 2 Proportion | 0.133079432 | Interval Lower Limit | -0.0854 | |
Difference in Two Proportions | -0.08350632 | Interval Upper Limit | -0.0817 | |
Average Proportion | 0.0613 | |||
Z Test Statistic | -106.9171 | |||
Two-Tail Test | ||||
Lower Critical Value | -1.9600 | |||
Upper Critical Value | 1.9600 | |||
p-Value | 0.0000 | |||
Reject the null hypothesis |
we will conclude that the difference in the mortality rates in US is lower than UK becasue confidence interval contain only negative numbers the 92% confidence interval in the true difference in the mortality rates between the two countries is (-0.0854 ,-0.0817)
question 5 5. (a) A sample of 12 of bags of Calbie Chips were weighed (to...
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