{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solutions8

# solutions8 - ISDS Homework 8 Solutions Two Sample...

This preview shows pages 1–6. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ISDS Homework 8 - Solutions Two Sample Hypothesis Testing Panayiotis Skordi 1. Test the hypothesis that the average SAT math scores from students in California and New York are different. A sample of 45 students from California had an average score of 552, whereas a sample of 38 New York students had an average score of 530. Assume the population standard deviations for California and New York students are 105 and 114 respectively. a. Test at the = 0.05 level. b. Find the p-value for these samples. Answers a. 2 1 : = H 2 1 1 : H Where: 1 = the mean SAT scores of California students. 2 = the mean SAT scores of New York students. We can also express the hypothesis as: : 2 1 =- H : 2 1 1 - H We are given the following information: n X Californian Students 45 552 105 New York Students 38 530 114 Lets test this hypothesis at the = 0.05 level. We know that we have a two-tailed test since the alternative hypothesis is : 2 1 1 - H . Since 1 and 2 are known, and the sample sizes taken are both greater than 29, we can use the following formulas. Then we can find 96 . 1 % 5 . 2 2 = = Z Z from the normal tables. 1 We use the equation: 2 2 2 1 2 1 2 1 n n x x + =- 228 . 24 38 114 45 105 2 2 2 1 = + =- x x We can now calculate the test statistic, for this two sample test, in standardized form using: 2 1 ) ( ) ( 2 1 2 1 x x H X X Z---- = For this example: 908 . 228 . 24 ) 530 552 ( =-- = Z Do NOT reject Ho Reject Ho 1.96 0.908 Reject Ho 1.96 z Here we do not reject H . We conclude that there is not enough evidence to support a difference between the two states. b. Find the p-value for these samples. 2 Do NOT reject Ho 0.908 0.908 31.86% 18.14% 18.14% z The p-value is the sum of the green areas i.e. 18.14% + 18.14% = 36.28% 3 2. A company tracks satisfaction scores based on customer feedback from individual stores on a scale from 0 to 100. The following data represents the customer scores from Store 1 and Store 2. Store1 : 90 87 93 75 88 96 90 82 95 97 78 sample mean = 88.3, sample standard deviation=7.30 Store 2: 82 85 90 74 80 89 75 81 93 75 sample mean = 82.4, sample standard deviation=6.74 Assume population standard deviations are equal but unknown and that the population is normally distributed. Test the hypotheses H : 1 = 2 vs. H 1 : 1 P 2 H : 1 2 vs. H 1 : 1 &amp;gt; 2 H : 1 2 vs. H 1 : 1 &amp;lt; 2 Using = 0.10. 4 a. 2 1 : = H 2 1 1 : H Where: 1 = the mean satisfaction scores of store 1. 2 = the mean satisfaction scores of store 2. We can also express the hypothesis as: : 2 1 =- H : 2 1 1 - H We are given the following information: n X s Store 1 11 88.3 7.30 Store 2 10 82.4 6.74 Lets test this hypothesis at the = 0.10 level....
View Full Document

{[ snackBarMessage ]}

### Page1 / 16

solutions8 - ISDS Homework 8 Solutions Two Sample...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online