Question

You wish to test the following claim (H.) at a significance level of a 0.0 You obtain the following two samples of data Sample #1 Sample #2 376 88.9 53.1 37.6 56.1 102.4 74.2 43.1 51.7 60.3 41.4 72.2 61 52.3 60.8 48.5 59.1 63.4 59.8 44 58.4 57.2 56.3 52.6 46.3 49.2 36. 575 73 56.1 47.2 55.1 62.3 63.4 47.7 60.1 0.2 56.8 56.8 59.6 60.3 51.7 55.47.7 63.1 60, 54. 584 83.3 32.7 584 84. 1024 72 83.3 575 73.3 61.8 49.6 90.I 52 70.7 658 622 794 83.3 64.6 53.7 45 47.5 53.7 59.3 67 68.6 87.7 662 64.2 76.6 79.4 61.4 86.7 79.3 56.5 What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic What is the p-value for this sample? For this calculation, use the degrees of freedom reported from the technology you are using. (Report answer accurate to four decimal places.) p-value The p-value is.. Oless than (or equal to) a Ogreater than a This test statistie leads to a decision to... Oreject the nuall Oaccept the null Ofail to reject the null As such, the final conclusion is that... OThere is sufficient evidence to warrant rejection of the claim that the first OThere is not sufficient evidence to warant rejection of the claim that the ○The sample data support the claim that the first population mean is OThere is not sufficient sample evidence to support the claim that the first population mean is greater than the second population mean. first population mean is greater than the second population mean. greater than the second population mean. population mean is greater than the second population mean.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁< u₂
Alternative hypothesis: u₁ > u₂

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 3.015544
DF = 79

t = [ (x₁ - x₂) - d ] / SE

a) t = 3.32

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of 3.32.

b) Therefore, the P-value in this analysis is 0.001.

c) Interpret results. Since the P-value (0.001) is less than the significance level (0.05), we have to reject the null hypothesis.

d) Reject the null hypothesis.

e) There is sufficient evidence to warrant rejection of the claim that the first population mean is greater than the second population mean.