Question

A random sample of 173 university students was selected, and both the email service provider and cell phone provider were determined for each one, resulting in the accompanying data. State and test the appropriate hypotheses. (Test whether cell phone provider has an effect on email provider.) Cell Phone Provider ATT Verizon Other gmail 26 16 Email Provider Yahoo 30 27 10 Other 28 18 10 State the appropriate hypotheses. O Ho: PA* P2Pg for i = 1,2,3 Hy: at least one P, -p, for i = 1, 2, 3 and - 1,2,3 OH: PA-P2- Po fori - 1,2,3 Hy no p,s are equal O Ho: PA-PA - Pafori - 1,2,3 He at least one p, P, fori - 1, 2, 3 and j = 1, 2, 3 OHO PAPaPa for i=1,2,3 Hall p 's are equal Compute the test statistic value. (Round your answer to three decimal places.) What can we say about the p-value? P value < 0.005 0.005 < P-value < 0.01 O 0.01 < P-value < 0.025 0.025 < P value < 0.05 0.05 <P value < 0.10 Show all HAGUTANA NOTE

Answer 1

null and alternative hypothesis are

Ho: Pa-P2 - Pa for i = 1, 2, 3 Hat least one PP for i - 1, 2, 3 and j = 1, 2, 3

The expected values are calculated using the formula: (row total * column total)/grand total.

Expected Values Column 1 Column 2 Column 3 Total Row 1 84 x 50 173 = 24.277 61 x 50 173 = 17.63 28x50 = 8.092 50 173 Row 2 84 x 67 173 = 32.532 61 x 67 173 = 23.624 28 x 67 173 10.844 67 Row 3 84x56 173 = 27.191 61 x 56 173 19.746 28 x 56 173 = 9.064 56 Total 84 61 28 173

Squared Distances Column 1 Column 2 Column 3 Row 1 (26–24.277) 24.277 0.122 (16-17.63) 17.63 = 0.151 (8–8.092) 8.092 = 0.001 Row 2 (30–32.532) 0.197 (27–23.624) 23.624 0.482 (10-10.844) 10.844 -0.066 32.532 Row 3 (28–27.191) 27.191 = 0.024 (18-19.746) 19.746 = 0.154 (10-9.0642 9.064 = 0.097

The Chi-Squared statistic is computed as follows: n x? = (Oi; – E;;) Eij = 0.122 + 0.197 + 0.024 + 0.151 + 0.482 +0.154 + 0.001 +0.066 + 0.097 = 1.294 i=1

The corresponding p-value for the test is p = Pr(xỉ > 1.294) = 0.8624.

p value > 0.1