1. A machine produces 3-inch nails. A sample of 12 nails was selected and their lengths determined. The results are as follows: 2.97 2.86 2.82 2.86 2.95 2.80 2.81 2.98 2.88 2.83 2.95 2.95 Assuming that ? = 0.01, test the hypothesis that the population mean is equal to 3. •?State the null and alternate hypotheses •?Calculate the mean and standard deviation •?Determine which test statistic applies, and calculate it •?Determine the critical value(s). •?State your decision: Should the null hypothesis be rejected?
a) Here the null hypothesis is Ho: μ = 3 and the alternative hypothesis is Ha: μ ≠ 3.
b) The mean, = 2.8883 and the standard deviation, s = 0.0675
c) The test statistic for testing Ho is given by
= (2.8883 – 3)/(0.0675/√12) = - 5.728
d) Since a = 0.01, from Student’s t distribution with (n-1) = 11 degrees of freedom the critical value is given by,
Critical value = ±3.106
e) The decision rule is Reject Ho if |t| > 3.106
That is, we Reject Ho if t < -3.106 or t > 3.106
Here, |t| = 5.728 > 3.106
So we reject the null hypothesis Ho.
2. A sample of size n = 20 is selected from a normal population to construct a 90% confidence interval estimate for a population mean. The interval was computed to be (7.60 to 10.70). Determine the sample standard deviation.
The 90% confidence interval for a population mean is given by
(-, +),
where n = 20, = 1.729 [From Student’s t distribution with (n-1) =19 d.f.].
Now, the width of the interval is given by
(+) - (-) = 2
Thus, 2 = 10.70 – 7.60 = 3.1
That is, 2*1.729 *s/√20 = 3.1
That is, s = (3.1*√20)/(2*1.729) = 4.0091
Thus the standard deviation is s = 4.0091
3. A random sample of 41 observations was selected from a normally distributed population. The sample mean was ¯x = 60, and the sample variance was s2 = 21.0. Does the sample show sufficient reason to conclude that the population standard deviation is not equal to 6 at the 0.05 level of significance? Use the p-value method. •?State the null and alternate hypotheses •?Determine which test statistic applies, and calculate it •?Determine the corresponding probability, and compare to •?State your decision: Should the null hypothesis be rejected?
a) Here the null hypothesis is Ho: σ = 6 and the alternative hypothesis is Ha: σ ≠ 6.
b) The test statistic for testing Ho is given by,
, follows a Chi-square distribution with (n-1) = 40 d.f.
Here it is given that, n = 41, s2 = 21.0.
Thus the test statistic is given by,
= (41 -1)*21/36 = 23.3333
c) The p-value is given by
p-value = P[ 23.3333] = 0.9836
The decision rule is Reject Ho if the p-value < 0.05
Here, p-value = 0.9836 > 0.05.
So we fail to reject Ho.
Thus the sample does not show sufficient reason to conclude that the population standard deviation is not equal to 6 at the 0.05 level of significance.
4. An insurance company states that 75% of its claims are settled within 5 weeks. A consumer group selected a random sample of 65 of the company’s claims and found 43 of the claims were settled within 5 weeks. Is there enough evidence to support the consumer groups claim that fewer than 75% of the claims were settled within 5 weeks? Test using the traditional approach with a = 0.05. •?State the null and alternate hypotheses •?Calculate the sample proportion •?Determine which test statistic applies, and calculate it •?Determine the critical value(s). •?State your decision: Should the null hypothesis be rejected?
a) Here the null hypothesis is Ho: p = 0.75 and the alternative hypothesis is Ha: p < 0.75.
b) The sample proportion, = 43/65 = 0.6615
c) The test statistic for testing Ho is given by,
follows a Standard Normal distribution.
Here, n = 65.
Thus the test statistic is given by
z = (0.6615 – 0.75)/Sqrt[0.75*0.25/65] = -1.6471
d) The critical value = -1.645
e) The decision rule is Reject Ho if z < -1.645
Here, z = -1.6471 < -1.645
So we reject Ho.Thus there is enough evidence to support the consumer group’s claim that fewer than 75% of the claims were settled within 5 weeks
5. A teacher wishes to compare two different groups of students with respect to their mean time to complete a standardized test. The time required is determined for each group. The data summary is given below. Test the claim at = 0.10, that there is no difference in variance. Give the critical region, test statistic value, and conclusion for the F test. n1 = 40 s1 = 34 n2 = 60 s2 = 41 ? = 0.10 •?State the null and alternate hypotheses •?Determine which test statistic applies, and calculate it •?Determine the critical region •?State your decision: Should the null hypothesis be rejected?
a) Here the null hypothesis is Ho: and the alternative hypothesis is Ha: .
b) The test statistic for testing Ho is given by,
, follows F distribution with ()d.f.
Here it is given that, n1 = 40, s1 = 34, n2 = 60, s2 = 41.
Thus the test statistic is given by,
F = (412/342) = 1.4542
c) Since a = 0.10, from F distribution with (59, 39) degrees of freedom the critical value is obtained as 1.474
Thus the critical region is F > 1.474
Here F = 1.4542 < 1.474
So we fail to reject the null hypothesis Ho. Thus there is no significant difference in variance.
6. A machine produces 9 inch latex gloves. A sample of 85 gloves is selected, and it is found that 36 are shorter than they should be. Find the 98% confidence interval on the proportion of all such gloves that are shorter than 9 inches.
The 99% confidence interval for the proportion is given by,
,
where = 2.326.
Here it is given that, n = 85, x = 36
Therefore, the sample proportion, = x/n = 36/85 = 0.4235
Thus, the 98% confidence interval on the proportion of all such gloves that are shorter than 9 inches is given by
(0.4235 – 2.326*√(0.4235*0.5765/85), 0.4235 + 2.326*√(0.4235*0.5765/85))
= (0.4235 – 0.1247, 0.4235 + 0.1247)
= (0.2988, 0.5482)
7. The pulse rates below were recorded over a 30-second time period, both before and after a physical fitness regimen. The data is shown below for 8 randomly selected participants. Is there sufficient evidence to conclude that a significant amount of improvement took place? Assume pulse rates are normally distributed. Test using ? = 0.10. •?State the null and alternate hypotheses •?Calculate the mean and standard deviation •?Determine which test statistic applies, and calculate it •?Determine the critical value(s). •?State your decision: Should the null hypothesis be rejected? Before 45 33 30 55 33 32 57 49 After 49 33 35 61 41 32 57 56
Let μ1 and μ2 denote the mean pulse rate before and after a physical fitness regimen respectively
Here the null hypothesis is Ho: μ1 = μ2 and the alternative hypothesis is Ha: μ1 < μ2.
The mean and standard deviation of the pulse rate before physical fitness regimen is
Mean = 41.75 and standard deviation = 11.068
The mean and standard deviation of the pulse rate after physical fitness regimen is
Mean = 45.5 and standard deviation = 11.7352
The mean and standard deviation of the difference in pulse rate before and after physical fitness regimen is
Mean = -3.75 and standard deviation = 3.3274
For testing Ho, we use the paired t test.
The test statistic for testing Ho is given by,
t = dbar/(sd/√n), follows a student’s t distribution with (n-1) degrees of freedom.
Here,dbar = -3.75, sd = 3.3274 and n = 8.
Thus the test statistic is given by,
t = -3.75/(3.3274/√8) = -3.1877
Since a = 0.10, from Student’s t distribution with (n-1) = 7 degrees of freedom the critical value is given by,
Critical value = -1.415
The decision rule is Reject Ho if t < -1.415
Here, t = -3.1877 <-1.415
So we reject the null hypothesis Ho.
Thus there is sufficient evidence to conclude that a significant amount of improvement took place at 5% significance level.
8. You are given the following data. Test the claim that there is a difference in the means of the two groups. Use ? = 0.01. Group A Group B ¯x1= 2 ¯x 2 = 8 s_1= 17 s_2 = 12 n_1 = 44 n_2 = 95 •?State the null and alternate hypotheses •?Determine which test statistic applies, and calculate it •?Determine the critical value(s). •?State your decision: Should the null hypothesis be rejected?
Here the null hypothesis is Ho: μ1 = μ2 and the alternative hypothesis is Ha: μ1 ≠ μ2.
Since the sample sizes are large we can use the z test for testing Ho.
The test statistic for testing Ho is given by,
t = (xbar1 – xbar2)/Sqrt[(s1^2)/n1+(s2^2)/n2] follows a Standard Normal distribution(approximately).
Here it is given that,xbar1= 2, xbar2 = 8, s1=17 , s2= 12, n1= 44, n2= 95,
Thus the test statistic is given by,
t = (2 – 8)/Sqrt[(17^2)/44+(12^2)/95] = - 2.1103
Since a = 0.01, from Standard Normal distribution the critical value is given by,
Critical value = ±2.58
The decision rule is Reject Ho if |t| > 2.58
or Reject Ho if t < -2.58 or t > 2.58.
Here, |t| = 2.1103 2.58
So we fail to reject the null hypothesis Ho.Thus there is no difference in the means of the two groups