Two-sample confidence interval difference of means

3 An engineering company buys a certain type of component from two suppliers, A and B. It is important that, on the whole, the strengths of these components are the same from both suppliers. The company can measure the strengths in its laboratory. Random samples of seven components from supplier A and five from supplier B give the following strengths, in a convenient unit. The underlying distributions of strengths are assumed to be Normal for both suppliers, with variances 2.45 for supplier A and 1.40 for supplier B.

1. Test at the \(5 \%\) level of significance whether it is reasonable to assume that the mean strengths from the two suppliers are equal.
2. Provide a two-sided 90\% confidence interval for the true mean difference.
3. Show that the test procedure used in part (i), with samples of sizes 7 and 5 and a \(5 \%\) significance level, leads to acceptance of the null hypothesis of equal means if \(- 1.556 < \bar { x } - \bar { y } < 1.556\), where \(\bar { x }\) and \(\bar { y }\) are the observed sample means from suppliers A and B . Hence find the probability of a Type II error for this test procedure if in fact the true mean strength from supplier A is 2.0 units more than that from supplier B.
4. A manager suggests that the Wilcoxon rank sum test should be used instead, comparing the median strengths for the samples of sizes 7 and 5 . Give one reason why this suggestion might be sensible and two why it might not.

10 The times taken to run 400 metres by students at two large colleges \(P\) and \(Q\) are being compared. There is no evidence that the population variances are equal. The time taken by a student at college \(P\) and the time taken by a student at college \(Q\) are denoted by \(x\) seconds and \(y\) seconds respectively. A random sample of 50 students from college \(P\) and a random sample of 60 students from college \(Q\) give the following summarised data. $$\Sigma x = 2620 \quad \Sigma x ^ { 2 } = 138200 \quad \Sigma y = 3060 \quad \Sigma y ^ { 2 } = 157000$$

1. Using a 10\% significance level, test whether, on average, students from college \(P\) take longer to run 400 metres than students from college \(Q\).
2. Find a \(90 \%\) confidence interval for the difference in the mean times taken to run 400 metres by students from colleges \(P\) and \(Q\).

9 The leaves from oak trees growing in two different areas \(A\) and \(B\) are being measured. The lengths, in cm , of a random sample of 7 oak leaves from area \(A\) are $$6.2 , \quad 8.3 , \quad 7.8 , \quad 9.3 , \quad 10.2 , \quad 8.4 , \quad 7.2$$ Assuming that the distribution is normal, find a 95\% confidence interval for the mean length of oak leaves from area \(A\). The lengths, in cm, of a random sample of 5 oak leaves from area \(B\) are $$5.9 , \quad 7.4 , \quad 6.8 , \quad 8.2 , \quad 8.7$$ Making suitable assumptions, which should be stated, test, at the \(5 \%\) significance level, whether the mean length of oak leaves from area \(A\) is greater than the mean length of oak leaves from area \(B\). [9]

Fish of a certain species live in two separate lakes, \(A\) and \(B\). A zoologist claims that the mean length of fish in \(A\) is greater than the mean length of fish in \(B\). To test his claim, he catches a random sample of 8 fish from \(A\) and a random sample of 6 fish from \(B\). The lengths of the 8 fish from \(A\), in appropriate units, are as follows. $$\begin{array} { l l l l l l l l } 15.3 & 12.0 & 15.1 & 11.2 & 14.4 & 13.8 & 12.4 & 11.8 \end{array}$$ Assuming a normal distribution, find a \(95 \%\) confidence interval for the mean length of fish in \(A\). The lengths of the 6 fish from \(B\), in the same units, are as follows. $$\begin{array} { l l l l l l } 15.0 & 10.7 & 13.6 & 12.4 & 11.6 & 12.6 \end{array}$$ Stating any assumptions that you make, test at the \(5 \%\) significance level whether the mean length of fish in \(A\) is greater than the mean length of fish in \(B\). Calculate a 95\% confidence interval for the difference in the mean lengths of fish from \(A\) and from \(B\).

1. The weights of the contents of breakfast cereal boxes are normally distributed.

A manufacturer changes the style of the boxes but claims that the weight of the contents remains the same.
A random sample of 6 old style boxes had contents with the following weights (in grams). $$\begin{array} { l l l l l l } 512 & 503 & 514 & 506 & 509 & 515 \end{array}$$ The weights, \(y\) grams, of the contents of an independent random sample of 5 new style boxes gave $$\bar { y } = 504.8 \text { and } s _ { y } = 3.420$$

1. Use a two-tail test to show, at the \(10 \%\) level of significance, that the variances of the weights of the contents of the old and new style boxes can be assumed to be equal. State your hypotheses clearly.
2. Showing your working clearly, find a \(90 \%\) confidence interval for \(\mu _ { x } - \mu _ { y }\), where \(\mu _ { x }\) and \(\mu _ { y }\) are the mean weights of the contents of old and new style boxes respectively.
3. With reference to your confidence interval comment on the manufacturer's claim.

The weights of the contents of breakfast cereal boxes are normally distributed. A manufacturer changes the style of the boxes but claims that the weight of the contents remains the same. A random sample of 6 old style boxes had contents with the following weights (in grams). 512, 503, 514, 506, 509, 515 The weights, \(y\) grams, of the contents of an independent random sample of 5 new style boxes gave $$\bar{y} = 504.8 \text{ and } s_y = 3.420$$

1. Use a two-tail test to show, at the 10\% level of significance, that the variances of the weights of the contents of the old and new style boxes can be assumed to be equal. State your hypotheses clearly. [5]
2. Showing your working clearly, find a 90\% confidence interval for \(\mu_x - \mu_y\) where \(\mu_x\) and \(\mu_y\) are the mean weights of the contents of old and new style boxes respectively. [7]
3. With reference to your confidence interval comment on the manufacturer's claim. [2]