Question

I ONLY NEED HELP WITH PART OF PART "B"

I've figured out the test statistic is -1.73 and the degrees of freedom are 5. However, I'm having a hard time finding the P value via the chart (which I'm required to learn how to do).I think the chart immediately bellow this is the one used to find the p-value. However, I know at least one (or more) of the charts bellow is what's used. Please let me know which chart is the one used to find the p-value (using the t-value) and how it can be done from its usage (or of there are multiple from their usages).

Answer 1

Answer a)

For doing hypothesis testing first we need to calculate mean and standard deviation of data provided:

To find standard deviation we use the following formula Tn-1 We will compute this formula in 4 steps. Step 1: Find the mean, X

x̅ = (400+500+600+600+700+800)/6 = 3600/6 = 600

Step 2: Create the following table. data 400 500 600 600 700 800 data-mean 200 100 (data-mean 40000 10000 100 10000 40000 200 Step 3: Find the sum of numbers in the last column to get. -X100000 Step 4: Calculate σ using the above formula. 141.4214 6-1

s = 141.4214

Answer b)

The provided sample mean is X = 600 and the sample standard deviation is s = 141.4214, and the sample size is n = 6. The following null and alternative hypotheses need to be tested: Но 2700 Ha: u < 700 This coresponas to aketes tor n mean wunknown popuation standard deviation will be used. 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

that the population mean μ is less than 700, at the 0.1 significance level.

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)

t Table cum. prob 90 975 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-tails 1.00 0.50 0.400.30 0.20 0.10 0.05 0.02 0.01 0.002 0.001 10.000 1.000 1.376 1.963 3.078 6.314 12.7 31.82 63.66 318.31 636.62 0.000 0.816 106 1.386 1.886 .920 4.3036.965 9.925 22.327 31.599 3 0.000 0.765 0.978 250 1.638 2.353 3.182 4.54 5.84 0.215 12.924 1.533 2.132 2.776 3.747 4.604 7.173 8.610 5 0.000 0.727 0.920 1.156 1476 2.015 2.573.365 4.032 5.893 6.869 1.134 1.440 1.943 2.447 3.143 3.707 5.208 5.959 7 0.000 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 4 0.000 0.741 0.941 1.190 6 0.000 0.718 0.906 91 0.000 0.703 0.883 1.100 1.383 1.833 2.262 2.821 3.250 4.297 4.781 10 0.000 0.700 0.879 1.093 1.372 1.812 2.228 .764 3.169 4.144 4.587 0.000 0.697 0.876 .088 1363 1.796 2.718 3.106 4.025 4.437 12 0.000 0.695 0.873 1.083 1.356 1.782 2.179 2.681 3.055 3.9304.318 13 0.000 0.694 0.870 1.079 1.350 1.77 2.160 2.650 3.012 3.852 4.221 14 0.000 0.692 0.868 1.076 1345 1.761 2.145 2.624 2.977 3.787 4.140 15 0.000 0.691 0.866 .074 1.341 1.753 2.131 2.602 2.947 3.733 4.073 16 0.000 0.690 0.865 1.071 1337 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 2.110 2.567 2.898 3.646 3.965 18 0.000 0.688 0.862 .067 1.330 1.734 2.101 2.552 2.878 3.610 3.922 19 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 1.323 172 2.080 2.518 2.831 3.527 3.819 22 0.000 0.686 0.858 1.061 1.321 1.717 2.074 2508 2.819 3.505 3.792 23 0.000 0.685 0.858 1.060 1.319 1.714 2.069 2.500 2.807 3.485 3.768 24 0.000 0.685 0.857 1.059 1.318 1.711 2.064 2.492 2.797 3.467 3.745 25 0.000 0.684 0.856 1.058 1.316 1708 2.060 2.485 2.787 3.450 3.725 26 0.000 0.684 0.856 1.058 1.315 1.706 2.056 2.479 2.779 3.435 3.707 27 0.000 0.684 0.855 .057 1.314 1.703 2.052 2.473 2.771 3.4213.690 28 0.000 0.683 0.855 1.056 1.313 1.701 2.048 2.467 2.763 3.408 3.674

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)

t Table cum. 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

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.