Question

36) Following are observed frequencies. The null hypothesis is $H_0: p_1 = 0.3, p_2 = 0.3, p_3 = 0.2,$ $p_4 = 0.05, p_5 = 0.015.$ Category | 1 | 2 | 3 | 4 | 5 ------- | -------- | -------- | -------- | -------- | -------- Observed | 23 | 35 | 16 | 19 | 12 Compute the value of $\chi^2$. ID: BESB80 11.1.3-2+

Name: 36 following are observed frequencies the null hypothesis is h0 p1 03 p2 03 p3 02 p4 005 p5 0015 category 1 2 3 4 5 observed 23 35 16 19 12 compute the value of chi2 id besb80 1113 2 22415
Uploaded: 2024-07-31T07:35:00-08:00
Duration: 1 min 8 s
Channel: Adi S
Description: 36 following are observed frequencies the null hypothesis is h0 p1 03 p2 03 p3 02 p4 005 p5 0015 category 1 2 3 4 5 observed 23 35 16 19 12 compute the value of chi2 id besb80 1113 2 22415

          36) Following are observed frequencies. The null hypothesis is $H_0: p_1 = 0.3, p_2 = 0.3, p_3 = 0.2,$
$p_4 = 0.05, p_5 = 0.015.$
Category | 1 | 2 | 3 | 4 | 5
------- | -------- | -------- | -------- | -------- | --------
Observed | 23 | 35 | 16 | 19 | 12
Compute the value of $\chi^2$.
ID: BESB80 11.1.3-2+

36 following are observed frequencies the null hypothesis is h0 p1 03 p2 03 p3 02 p4 005 p5 0015 category 1 2 3 4 5 observed 23 35 16 19 12 compute the value of chi2 id besb80 1113 2 22415

Added by Erika J.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Adi S

07/31/2024

Step 1

3) = 33$ $E_2 = np_2 = 110(0.3) = 33$ $E_3 = np_3 = 110(0.2) = 22$ $E_4 = np_4 = 110(0.05) = 5.5$ $E_5 = np_5 = 110(0.015) = 1.65$ Show more…

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36) Following are observed frequencies. The null hypothesis is $H_0: p_1 = 0.3, p_2 = 0.3, p_3 = 0.2,$ $p_4 = 0.05, p_5 = 0.015.$ Category | 1 | 2 | 3 | 4 | 5 ------- | -------- | -------- | -------- | -------- | -------- Observed | 23 | 35 | 16 | 19 | 12 Compute the value of $\chi^2$. ID: BESB80 11.1.3-2+