Question 1: - JustAnswer



Question 1:

Let p1 = p Con, p2 = p BKG, n1 = n Con, and n2 = n BKG then can test Ho: p1 = p2 vs Ha; p1 > p2 + .02 with the normal approximation Z statistic given by

Z = (p1 – p2 – .02) / SQRT[p1*(1–p1) / n1 + p2*(1–p2) / n2].

For alpha = 0.05 reject Ho if Z > 1.645 and for alpha = 0.01 reject Ho if

Z > 2.327.

Z = ( 0.2 – 0.14 - 0.02) / SQRT[.2*.8 / 350 + .14*.86 / 400]

= .04 / SQRT(.000457 + .000301)

= .04 / .02753

= 1.453

At neither alpha = 0.05 or alpha = 0.01 can Ho be rejected. Can not conclude that CON profit is more than 2 % greater than BKG.

Question 2:

Before After Diff(B – A)

5.9 5.4 0.5

7.5 7.1 0.4

6.1 6.2 –0.1

6.8 6.5 0.3

8.1 7.8 0.3

Ho: Diff = 0. Ha: Diff > 0. One-sided test. Test using T-test which is

T = (sample mean – 0) / (sd / √n) with 4 degrees of freedom. Reject at

alpha = 0.01 level if T > 3.75

T = 0.28 / (.228 / √5)

= 0.28 / .102

= 2.745

Can not reject Ho since 2.745 < 3.75. Conclusion is that there has not been a significant decrease in time.

Question 3:

Ho: u1 = u2, Ha: u1 > u2. This is a one tailed test.

When sample sizes are equal can estimate the pooled variance, S^2 by mean of two sample variances. s1 = 3.3 so s1^2 = 10.89. s2 = 2.1, so s2^2 = 4.41. Mean is (10.89 + 4.41) / 2 = 7.65. Then use T statistic

T = (SM1 – SM2) / SQRT{S^2*[(n1 + n2) / (n1*n2)]}

where SM1 is group 1 sample mean, SM2 is group 2 sample mean, S^2 is the pooled estimate of variance, n1 is the group 1 sample size, n2 is group 2 sample size and SQRT is the square root. T-statistic has 10 (n1 + n2 – 2) degrees of freedom so can reject Ho if T > 1.812 (from T table or T calculator). So,

T = (75 – 72) / SQRT[7.65*(12 / 36)]

= 3 / 1.597

= 1.879

T = 1.879 > 1.812 so Ho can be rejected.

Question 4:

A’s B’s C’s D’s Rest Total

Observed 10 20 20 5 5 60

Expected 6 12 18 12 12 60

The expected are computed from the percentages from the past record.

To test the hypothesis, Ho, that Dr. Johnson’s class has the same distribution as the historical grade distribution, the chi-square statistic will be used. It is

Chi–square = SUM {(O – E)^2 / E)

where the sum is over the 5 cells in the table. The Chi-square has 4 degrees of freedom. At alpha = 0.05 can reject Ho if Chi-square > 9.49 and for alpha = 0.01 can reject Ho if Chi-square > 13.28 (from chi-square table or calculator).

So,

Chi-square = (10 – 6)^2 / 6 + (20 – 12)^2 / 12 + (20 – 18)^2 / 18

+ (5 – 12)^2 / 12 + (5 – 12)^2 / 12

= 2.667 + 5.333 + .222 + 4.083 + 4.083

= 16.388

Ho can be rejected at both alpha = 0.05 and alpha = 0.01 levels. Conclusion is that Dr. Johnson’s class did not have same distribution as historical distribution.

Question 5:

1 2 3 4 5 ≥6 Total

Observed 120 81 48 18 18 15 300

Expected 90 75 45 36 24 30 300

To test hypothesis, Ho, that small town distribution is the same as national distribution, use same procedure as in Question 4 but here Chi-square has 6 cells so it has 5 degreees of freedom. For alpha = 0.05 can reject Ho if chi-square > 11.07 and for alpha = 0.01 can reject Ho if Chi-square > 15.09. So,

Chi-square = (120 – 90)^2 / 90 + (81 – 75)^2 / 75 + (48 – 45)^2 / 45

+ (18 – 36)^2 / 36 + (18 – 24)^2 / 24 + (15 – 30)^2 / 30

= 10 + 0.48 + 0.20 + 9 + 1.5 + 7.5

= 28.68

Ho can be rejected at both alpha = 0.05 and alpha = 0.01 levels. Conclusion is that the admission distribution of the small town is different from the national distribution. Inspection of the table above shows in the small town there are more in the 1 admission category and fewer in the ≥6 category than expected from the national survey. Those facts together with the significant chi-square support the small town mayors claim that his city is healthier than national average.

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