1)In a study of the effectiveness of a new, lighter, running shoe you have desig

Question

1)In a study of the effectiveness of a new, lighter, running shoe you have designed, you give the shoe to 1068 runners, and compare them with a sample of 2127 runners who don’t receive the shoe. 698 of the sample of runners with the shoes won their race, compared to 639 runners in the control condition. Test the prediction that the shoe is significantly more effective in enhancing running performance at the alpha = .05 level.

. A)Because this data has two sets of contingencies (won versus not win), and (shoes versus no shoes), I can set up a 2 x 2 contingency table and use the Chi-square distribution.

B. The critical value of my statistic at 1 degree of freedom is 3.84

C). The computed value of my statistic is 364.35

D). I must reject the null hypothesis that the proportion of people who win with the shoe is the same as the proportion of people who win without the shoe.

E. All the above

2) The following measurements were made in inches, for the inside diameter of steel pipe to be used to make a Rally Car frame: 0.61, 0.62, 0.61, 0.58, 0.60, 0.61, 0.61, 0.61, 0.40, 0.51. For this sample, which of the following statements are true?

The 95% Confidence Interval of the Mean, for the sampling distribution for n= 10 observations is 0.54 inches to 0.62 inches.

B)The probability of drawing a .40 from a population with the same mean and standard deviation as this sample is about 0.005

C)To make the claim in b, you have to assume independence of each measurement.

D)A only

E)B only

F) B and C

G)A,b,c

A.

The 95% Confidence Interval of the Mean, for the sampling distribution for n= 10 observations is 0.54 inches to 0.62 inches.

B)The probability of drawing a .40 from a population with the same mean and standard deviation as this sample is about 0.005

C)To make the claim in b, you have to assume independence of each measurement.

D)A only

E)B only

F) B and C

G)A,b,c

Explanation / Answer

Q1.

Given table data is as below

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calculation formula for E table matrix

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expected frequecies calculated by applying E - table matrix formulae

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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above

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set up null vs alternative as

null, Ho: no relation b/w X and Y OR X and Y are independent

alternative, H1: exists a relation b/w X and Y OR X and Y are dependent

level of significance, = 0.05

from standard normal table, chi square value at right tailed, ^2 /2 =3.8415

since our test is right tailed,reject Ho when ^2 o > 3.8415

we use test statistic ^2 o = (Oi-Ei)^2/Ei

from the table , ^2 o = 364.3457

critical value

the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 3.8415

we got | ^2| =364.3457 & | ^2 | =3.8415

make decision

hence value of | ^2 o | > | ^2 | and here we reject Ho

^2 p_value =0

ANSWERS

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null, Ho: no relation b/w X and Y OR X and Y are independent

alternative, H1: exists a relation b/w X and Y OR X and Y are dependent

test statistic: 364.3457

critical value: 3.8415

p-value:0

decision: reject Ho

your asnwer should be, E. All the above

col1 col2 698 370 1068 639 1488 2127 1337 1858 N = 3195

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