Question

1. Suppose that you work in a shoe company want to compare two materials, A and B, for use on the soles of boys' shoes. In this problem, each of ten boys in a study wore a special pair of shoes with the sole of one shoe made from Material A in column (Mat-A) and the sole on the other shoe made from Material B in column (Mat-B). The sole types were randomly assigned to account for systematic differences in wear between the left and right foot. After three months, the shoes are measured for wear

data:

weight in lb	Mat-A	Mat-B
49	13.2	14
51	8.2	8.8
46	10.9	11.2
50	14.3	14.2
51	10.7	11.8
47	6.6	6.4
44	9.5	9.8
47	10.8	11.3
46	8.8	9.3

• • Weight measurements were made on nine boys in column (weight lb). You know that the distribution of measurements has historically been close to normal with s= 0.2. Test if the population mean is 50 and obtain a 90% confidence interval for the mean. - (Solve manually and Minitab)
• • You want to see if these is difference between the two materials. Justify your answers by using hypothesis testing and confidence interval procedures. - (Solve manually and Minitab)
  • Compare the results from the paired procedure with those from an unpaired- (Solve manually and Minitab)

Answer 1

a) To test H₀ :

The test statistic, t =  , has t distribution with (n-1) degrees of freedom

t =

At 10% significance level the critical value obtained from t table at 8 degrees of freedom is Since t lies outside the range (-1.86, + 1.86)

The hypothesis that is rejected

90% confidence interval =

= (47.88 - 1.6256 , 47.88 + 1.6256)

= ( 46.2543 , 4950)

b) To test H₀ :

the test statistic t =

Note : We should have two assumptions in place for using the above test statistic

1) The population from which the sample of Matrix A & B taken are normally distributed

2) The standard deviations are assumed to be equal. The second assumption is tested using F test and it can be assumed to be equal.

S₁ = 2.4021 , S₂ = 2.4951 ,

On subsitution

unpaired t = -0.5396

90% confidence interval is

The critical value is obtained from t table at 10% significance level and n₁ + n₂ - 2 = 16 degrees of freedom i.e 1.746 = t_0.90

Therefore 90% C 1 is = (-0.4226 - 1.746 * 0.78 , -0.4226 + 1.746 * 0.78)

= (-1.78 , 0.9393)

Since the 90% confidence interval contains 0. We accept the null hypothesis at 10% level of significance.

This conclusion can also be arrived using the critical level comparison with the test statistical value - 0.5396. The statistic is within the critical value range ( - 1.746 , +1.746 ). Hence we accept the null hypothesis at a 10% level of significance. There is no difference observed in the use of 2 materials of soles.

c) paired t-test :

To test H₀ :

The test statistics =

= - 2.92281

The critical value for 8 degrees of freedom, at a 10% significance level is 1.86.

On comparing the critical value with test statistic value we reject the null hypothesis i.e this hypothesis was accepted in unpaired t-test but was rejected in paired t-test. According to paired t-test, the difference is actually evident at 10% significance level.