Question

2.7. Suppose that we are testing Ho : μ1 Ha versus Ho: > μ2 where the two sample sizes are ni n, 10. Both sample variances are unknown but assumed equal. Find bounds on the P-value for the following observed values of the test statistic. (a) to= 2.31 (b) toー3.60 (c) to-1.95 (d) 0-219 2.8. Consider the following sample data: 9.37, 13.04, 11.69, 8.21, 11.18, 10.41, 13.15, 11.51, 13.21, and 7.75. Is it reasonable to assume that this data is a sample from a normal distribution? Is there evidence to support a claim that the mean of the population is 10? 2.9. put for a hypothesis-testing problem A computer program has produced the following out- Difference in sample means: 2.35 Degrees of freedom: 18 Standard error of the difference in sample means: ? Test statistic: to2.01 P-value: 0.0298 (a) What is the missing value for the standard error? (b) Is this a two-sided or a one-sided test? (c) If α 0.05, what are your conclusions? (d) Find a 90% two-sided CI on the difference in means

Please solve 2.7, 2.8 and 2.9 problems.

Answer 1

Solution:

    27) We are given that:

$H_{0} : \mu_{1} = \mu_{2}$ Vs $H_{1} : \mu_{1} > \mu_{2}$

Sample sizes are n1 = n2 = 10

Variances are unknown but assumed equal.
We have to find bounds on P-value for the following observed values of the test statistic.

Since $H_{1} : \mu_{1} > \mu_{2}$ is right tailed , we use one tail area for finding P-value intervals.

df = n1 + n2 - 2 = 10 + 10 - 2 = 18

So we look in t table for row of df = 18 and find the interval in which t test statistic fall.

Part a) t₀ = 2.31

Look in t table for df row = 18 and find the interval in which t = 2.31 fall , then find corresponding one tail area interval, which would be bounds on P-value.

t₀ = 2.31 fall in between t = 2.101 to t = 2.552, thus corresponding one tail area is in between 0.01 to 0.025

Thus bounds on P-value are:

0.01 < P-value < 0.025

Part b) t₀ = 3.60

Look in t table for df row = 18 and find the interval in which t = 3.60 fall , then find corresponding one tail area interval, which would be bounds on P-value.

t Table cum. prob one-tail0.50 0.25 0.20 0.15 0.10 0.05 0.025 0.00.005 .001 0.0005 two-tails1.00 0.50 0.40 0.30 0.20 0.10 0.05 0.02 0101 0.002 0.001 2 1 0.000 000 1.376 1.963 3.078 6.314 12.71 31.82 2 0.000 0.816 .06 386 1886 2.920 4.303 6.965 9.925 2227 31.599 3 0.000 0.765 0.978 1.250 638 2.353 3.182 4.545.41 10215 12.924 40.000 0.741 0.94 1.190 533 2.132 2.776 3.7474.04 7.73 8.610 5 0.000 0.727 0.920 .156 .476 2.015 2.571 3.365 4.32 5.893 6.869 6 0.000 0.718 0.906 1.134 1.440 1.943 2.447 3.143 3.107 5.108 5.959 71 0.000 0.711 0.896 1.119 1.415 1.895 2.365 2.998 3. 99 4.185 5.408 8 0.000 0.706 0.889 1.108 1.397 1.860 2.306 2.896 3.B55 4.501 5.041 9 0.000 0.703 0.883 1.100 1.383 1.833 2.262 2.821 3250 4.197 4.781 1 100.000 0.700 0.879 1.093 1.372 1.812 2.228 2.764 11 0.000 0.697 0.876 1.088 .363 1.796 2.202.718 120.000 0.695 0.873 .083 .356 1.782 2.179 2.681 13 0.000 0.694 0.870 .079 1.350 1.7712.160 2.650 14 0.000 0.692 0.868 1.076 .345 1.76 2.145 2.624 15 0.000 0.691 0.866 1.074 341 753 2.13 2.602 16 0.000 0.690 0.865 1.071 1.337 1.746 2.120 2.583 3

t₀ = 3.60 fall in between t = 2.878 to t = 3.610, thus corresponding one tail area is in between 0.001 to 0.005

Thus bounds on P-value are:

0.001 < P-value < 0.005

Part c) t₀ = 1.95

Look in t table for df row = 18 and find the interval in which t = 1.95 fall , then find corresponding one tail area interval, which would be bounds on P-value.

t Table cum. prob 2 one-tail0.50 0.25 0.20 0.15 0.10 0.05 0.025 0.0 0.005 0.0010.0005 2 1 0 2 0 1 0.000 000 1.376 1.963 3.078 6B14 12.11 31.82 63.66 318.31 636.62 2 0.000 0.816 .06 386 1886 2 920 4.33 6.965 9.925 22.327 31.599 3 0.000 0.765 0.978 1.250 638 2853 3.12 4.545.841 10.215 12.924 40.000 0.741 0.94 1.190 533 2132 2.776 3.747 4.604 7.173 8.610 5 0.000 0.727 0.920 .156 .476 2015 2.571 3.365 4.032 5.893 6.869 6 0.000 0.718 0.906 1.134 1.440 1.943 2.4473.143 3.707 5.208 5.959 71 0.000 0.711 0.896 1.119 1.415 1. 95 2 35 2.998 3.499 4.785 5.408 9 0.000 0.703 0.883 1.100 1.383 1.33 2.262 2.821 3.250 4.297 4.781 10 0.000 0.700 0.879 1.093 1.372 1.8122.228 2.764 3.169 4.144 4.587 111 0.000 0.697 0.876 1.088 1.363 6 2.718 3.106 4.025 4.437 12 0.000 0.695 0.873 1.083 1.356 1.7 2 2.179 2.68 3.055 3.930 4.318 13 0.000 0.694 0.870 .079 1.350 1.71 2.160 2.650 3.012 3.852 4.221 141 0.000 0.692 0.868 1.076 1.345 5 2.624 2.977 3.787 4.140 15 0.000 0.691 0.866 1.074 341 1753 2.131 2.602 2.947 3.733 4.073 16 0.000 0.690 0.865 1.071 1.337 1.746 2.120 2.583 2.921 3.686 4.015 171 0.000 0.689 0.863 1.069 1.333 1.740 2.10 2.567 2.898 3.646 3.965 18

t₀ = 1.95 fall in between t = 1.734 to t = 2.101, thus corresponding one tail area is in between 0.025 to 0.05

Thus bounds on P-value are:

0.025 < P-value < 0.05

Part d) t₀ = 2.19

Look in t table for df row = 18 and find the interval in which t = 2.19 fall , then find corresponding one tail area interval, which would be bounds on P-value.

t₀ = 2.19 fall in between t = 2.101 to t = 2.552, thus corresponding one tail area is in between 0.01 to 0.025

Thus bounds on P-value are:

0.01 < P-value < 0.025