Question

Question 1 of 4 For the following observed and expected frequencies: Observed 39 43 42 109 Expected 38 48 45 S 6 Download data Test the hypothesis that the distribution of the observed frequencies is as given by the expected frequencies. Use thea -0.025 level of significance and theP-value method with the TI-84 calculator Part 1 State the null and alternate hypotheses. Ho: The distribution of the observed frequencies ts H1: The distribution of the observed frequencies differs from that of the expected frequencies. that of the expected frequencies. Hi: The distribution of the observed frequencies differs from that of the expected frequencies. This hypothesis test is a right-tailed est. Part 2 Find the P-value. Round the answer to four decimal places. The P-value is Part 3 Determine whether to reject Ho (select) the null hypothesis Ho Part 4 State a conclusion. There (select) ▼ enough evidence to conclude that the distribution differs from what was expected. Following are observed frequencies. The null hypothesis is H0: P1-0.20,P2 = 0.25,P3-0.10,P4-0.30,p5-0.15 Category |12345 Observed 38 33 21 31 27 Download data Test the hypothesis that the distribution of the observed frequencies is as given by the null hypothesis. Use the 0.05 level of significance and thP -value method with the TI-84 calculator Part 1 State the null and alternate hypotheses. Ho: P1-0.20,P2-ㄈㄧㄧㄧ-P3-ㄈㄧ囗,P4-0.30,p5-0.15 This hypothesis test is a (select)test Part 2 Find the P-value. Round the answer to four decimal places. The P-value is□ー□. Part 3 Determine whether to reject Ho- the null hypothesis Ho- (select) Part 4 State a conclusion. enough evidence to conclude that the distribution differs from what was expected. There (select) Question 3 of 4 A statistics teacher claims that, on the average, 5% of her students get a grade of A, 15% get a B, 35% get a C, 30% get a D, and 15% get an F. The grades of a random sample of 120 students were recorded. The following table presents the results. Test the hypothesis that the grades follow the distribution claimed by the teacher. Use the 0.025 level of significance and tRe-value method with the TI-84 calculator. Grade A B D F Observed 9 448 37 22 Download data Part 1 How many of the students in the sample got a B? How many got a C? The number of students who got a Bi The number of students who got a C i Part 2 State the null and alternate hypotheses. H1: Some or all of the actual probabilities (select)from those specified byHo test This hypothesis test is a (select) Part 3 Find the P-value. Round the answer to four decimal places. The P-value is Part 4 Determine whether to reject Ho (select) wthe null hypothesis Ho Question 4 of 4 A governmental agency computed the proportion of violent crimes in the United States in a particular year falling into each of four categories. A simple random sample of 580 violent crimes committed in California during that year were categorized in the same way. The following table presents the results U.S. California Proportion Frequency Category Murder Forcible Rape Robbery Aggravated Assault0.590 0.015 0.055 0.340 35 202 339 Download data Can you conclude that the proportions of crimes in the various categories in California differ from those in the United States as a whole? Use the 0.01 level of significance and Revalue method with the TI-84 calculator Part 1 State the null and alternate hypotheses. Ho: The proportions of crimes in the various categories in Californial(select) the same as those in the United States as a whole. the same as those in Hi : The proportions of crimes in the various categories in California(select) the United States as a whole. This hypothesis test is a (select)test. Part 2 Find the P-value. Round the answer to four decimal places The P-value is Part 3 Determine whether to reject Ho Part 4 State a conclusion. There (select) enough evidence to conclude that the various categories in California differ from those in the United States as a whole pyright 2019 The McGraw-Hill Companies

Answer 1

Q1 1. Null and alternative hypothesis:

Ho: The distribution of the observed frequencies is equal to that of the expected frequencies

H1: The distribution of the observed frequencies differs from that of the expected frequencies

This is a right tailed test.

2. Test statistic:

χ² = ∑ ((O-E)²/E) = 0.0263 + 0.5208+0.2+0.5+1.5 = 2.7471

df = 5-1 =4

p-value = CHISQ.DIST.RT(2.7471, 4) =0.6010

3. Fail to reject the null hypothesis Ho.

4. There is not enough evidence to conclude.

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Q2: 1. Null and alternative hypothesis:

H0​:p1​=0.20,p2​=0.25,p3​=0.10,p4​=0.30,p5​=0.15

This is a right tailed test.

2.

The following table is obtained:

Categories	Observed	Expected	(fo-fe)2/fe
1	38	150*0.20=30	(38-30)2/30 = 2.133
2	33	150*0.25=37.5	(33-37.5)2/37.5 = 0.54
3	21	150*0.10=15	(21-15)2/15 = 2.4
4	31	150*0.30=45	(31-45)2/45 = 4.356
5	27	150*0.15=22.5	(27-22.5)2/22.5 = 0.9
Sum =	150	150	10.329

χ² = ∑ ((O-E)²/E) = 2.133+ 0.54+2.4+4.356+0.9 = 10.329

df = 5-1 =4

p-value = CHISQ.DIST.RT(10.329, 4) =0.0352

3. Reject the null hypothesis Ho.

4. There is enough evidence to conclude.

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Q3: 1. Number of student who get a B = 4

Number of student who got a C = 48

2. Null and alternative hypothesis:

H0​:p1​=0.05,p2​=0.15,p3​=0.35,p4​=0.30,p5​=0.15

Ha: Some or all of the actual probabilities differs from those specified by Ho.

This is a right tailed test.

3.

The following table is obtained:

Categories	Observed	Expected	(fo-fe)2/fe
A	9	120*0.05=6	(9-6)2/6 = 1.5
B	4	120*0.15=18	(4-18)2/18 = 10.889
C	48	120*0.35=42	(48-42)2/42 = 0.857
D	37	120*0.30=36	(37-36)2/36 = 0.028
F	22	120*0.15=18	(22-18)2/18 = 0.889
Sum =	120	120	14.163

χ² = ∑ ((O-E)²/E) = 14.163

df = 5-1 =4

p-value = CHISQ.DIST.RT(14.163, 4) = 0.0068

4. Reject the null hypothesis Ho.

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Q4: 1. Null and alternative hypothesis:

Ho: The proportions are the same.

H1: The proportions are not the same.

This is a right tailed test.

2. The following table is obtained:

Categories	Observed	Expected	(fo-fe)2/fe
Murder	4	580*0.015=8.7	(4-8.7)2/8.7 = 2.539
Rape	35	580*0.055=31.9	(35-31.9)2/31.9 = 0.301
Robbery	202	580*0.340=197.2	(202-197.2)2/197.2 = 0.117
Assault	339	580*0.590=342.2	(339-342.2)2/342.2 = 0.03
Sum =	580	580	2.987

χ² = ∑ ((O-E)²/E) = 2.987

df = 4-1 =3

p-value = CHISQ.DIST.RT(2.987, 3) = 0.3936

3. Fail to reject the null hypothesis Ho.

4. There is not enough evidence to conclude.