Answers
Answer a)
For doing hypothesis testing first we need to calculate mean and standard deviation of data provided:
x̅ = (400+500+600+600+700+800)/6 = 3600/6 = 600
s = 141.4214
Answer b)
Thus, it can be concluded that couple's sample average is less than true average not by pure confidence.
For your question related to p-value:
There are two ways to test hypothesis. One is critical value approach and other is p- value approach.
The result of the hypothesis testing will be same using both approaches.
Critical Value Approach:
In this approach, first we have to find out critical t value using the t-distribution table. Then we compare the critical t-value with the test statistics. If the magnitude of test statistic is greater than critical t-value, we reject null hypothesis.
In this case, to determine critical t value we need following:
Degree of freedom df = 6 -1 = 5
α = 0.10 (One tailed test)
Critical t value for α = 0.10 and df = 5 is 1.476 (Screenshot to show how we located t-value is attached)
In this case, since the magnitude of test statistic (1.732) is greater than critical t-value (1.476), so we reject null hypothesis
P-value Approach
Please not that using table, you cannot determine the exact p-value. You can only find out range.
In this case, for determining p-value we need two things:
Test statistics = 1.732
Degree of freedom df = 6 -1 = 5
P-value is between 0.10 < p < 0.05 (Screenshot to show how we find out range is attached)
In the t-distribution table, we need to see the row corresponding to df = 5. Now, our test statistics lies between 1.476 and 2.015. P-value (one tail) corresponding to 1.476 and 2.015 is 0.10 and 0.05 respectively. So, p-value lies between 0.05 and 0.10.
You can determine exact p-value using online calculator. The exact p-value is 0.0719 which is between 0.05 and 0.10.
In this case, since p value is less than α = 0.10 so we reject null hypothesis.
Step 1: Find the mean, X
0.1, and the critical value for a left- Based on the information provided, the significance level is α tailed test is te = 1.476. The rejection region for this left-tailed test is R-tt1.476 (3) Test Statistics The t-statistic is computed as follows X-110 600 700-1732 s/m 141.4214/6 (4) Decision about the null hypothesis Since it is observed thatt1.732te-1.476, it is then concluded that the null hypothesis is rejected Using the P-value approach: The p-value is p = 0.0719, and since p = 0.0719 that the null hypothesis is rejected. 0.1, it is concluded
Prolb 75 50 9 one-tail 0.50 0.25 0.20 0.15 0.10 0.05 0.025 0.01 0.005 0.001 0.0005 two-tails1.00 0.50 0.40 0.30 0.20 0.10 0.05 0.02 0.0 0.002 0.001 泭 10.000 1.000 376 1.963 3.078 6.314 12.71 31.82 63.66 318.31 636.62 2 0.000 0.816 1061 1386 1.886 2.920 4.303 6.965 9.925 22.327 31.599 3 0.000 0.765 0.978 1250 1.638 2.353 3.182 454 5810.215 12.924 4 0.000 0.741 0.941 1.190 1533 2.132 2.776 3.747 4.604 7.173 8.610 5 0.000 0.727 0.920 1.156 1.476 2.015 2.5 3.365 4.032 5893 6.869 6 0.000 0.718 0.906 1.134 1.440 1.943 2.447 3.143 3.707 5.208 5.959 70.000 0.711 0.896 1.119 1.415 1.895 2.365 2.998 3.499 4.785 5.408 8 0.000 0.706 0.889 1.108 1.397 1.860 2.306 2.896 3.355 4.501 5.041 9 0.000 0.703 0.883 1.100 1.383 1.833 2.262 2.821 3.250 4.297 4.781 100.000 0.700 0.879 1.093 1.372 1.812 2.228 2.764 3.169 4.144 4.587 11 0.000 0.697 0.876 1.088 1.363 796 2.201 2.718 3.106 4.025 4.437 12 0.000 0.695 0.873 1.083 1356 1.782 2.179 2.681 3.055 3.930 4.318 13 0.000 0.694 0.870 1.079 1.350 177 2.160 2.650 3.012 3.852 4.221 14 0.000 0.692 0.868 1.076 1.345 1761 2.145 2.624 2.977 3.7874.140 15 0.000 0.691 0.866 1.074 1.341 1.753 2.131 2.602 2.947 3.733 4.073 16 0.000 0.690 0.865 1.071 13371.746 2.120 2.583 2.921 3.686 4.015 17 0.000 0.689 0.863 1.069 1.333 1.740 2.110 2.567 2.898 3.646 3.965 18 0.000 0.688 0.862 1.067 1330 1.734 2.101 2.552 2.878 3.610 3.922 0.000 0.688 0.861 1.066 1.328 1.729 2.093 2.539 2.861 3.579 3.883 20 0.000 0.687 0.860 1.064 1.325 1.725 2.086 2.528 2.845 3.552 3.850 21 0.000 0.686 0.859 1.063 1.323 1.721 2.080 2.518 2.831 3.527 3.819 3