Consider the hypothesis test H0: μ1=μ2 against H1: μ1 ≠μ2 . Suppose that sample sizes are n1=15

Consider the hypothesis test H0: μ1=μ2 against H1: μ1 ≠μ2 ....

Consider the hypothesis test $H_{0}: \mu_{1}=\mu_{2}$ against $H_{1}: \mu_{1} \neq \mu_{2} .$ Suppose that sample sizes are $n_{1}=15$ and $n_{2}=15,$ that $\bar{x}_{1}=4.7$ and $\bar{x}_{2}=7.8,$ and that $s_{1}^{2}=4$ and $s_{2}^{2}=6.25 .$ Assume that $\sigma_{1}^{2}=\sigma_{2}^{2}$ and that the data are drawn from normal distributions. Use $\alpha=0.05$. (a) Test the hypothesis and find the $P$ -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) for a true difference in means of $3 ?$ (d) Assuming equal sample sizes, what sample size should be used to obtain $\beta=0.05$ if the true difference in means is $-2 ?$ Assume that $\alpha=0.05 .$

Consider the hypothesis test $H_{0}: \mu_{1}=\mu_{2}$ against $H_{1}: \mu_{1} \neq \mu_{2} .$ Suppose that sample sizes are $n_{1}=15$ and $n_{2}=15,$ that $\bar{x}_{1}=4.7$ and $\bar{x}_{2}=7.8,$ and that $s_{1}^{2}=4$ and $s_{2}^{2}=6.25 .$ Assume that $\sigma_{1}^{2}=\sigma_{2}^{2}$ and that the data are drawn from normal distributions. Use $\alpha=0.05$.
(a) Test the hypothesis and find the $P$ -value.
(b) Explain how the test could be conducted with a confidence interval.
(c) What is the power of the test in part (a) for a true difference in means of $3 ?$
(d) Assuming equal sample sizes, what sample size should be used to obtain $\beta=0.05$ if the true difference in means is $-2 ?$ Assume that $\alpha=0.05 .$

Key Concepts

P-Value

The P-value is the probability of observing test results at least as extreme as the ones computed from the sample data, assuming that the null hypothesis is true. It measures the strength of evidence against the null hypothesis, and a small P-value indicates that such extreme observed outcomes are unlikely under the null hypothesis, thereby supporting its rejection.

Sample Size Determination

Sample size determination involves calculating the number of observations required in a study to achieve a desired level of power for detecting a specified effect size at a given significance level. This concept is important in study design to ensure that the study is neither underpowered (risking a Type II error) nor unnecessarily large, optimizing resources while maintaining statistical validity.

Power Analysis

Power analysis is concerned with the probability of correctly rejecting a false null hypothesis, which is defined as the power of a test (1 - ?, where ? is the probability of a Type II error). It helps in evaluating how sensitive a test is to detect a true effect or difference and is crucial in interpreting the reliability of a test's result, especially when planning experiments to detect a specific effect size.

Two-Sample t-Test

The two-sample t-test is used when comparing the means of two independent samples to determine if there is a statistically significant difference between them. When the assumption of equal population variances is made, a pooled variance estimate is used to improve the accuracy of the test statistic. This test is fundamental in comparing two groups under the assumption of normally distributed data.

Hypothesis Testing

Hypothesis testing is a statistical inference method used to decide whether observed data are consistent with a stated claim or hypothesis about a population parameter. It involves setting up a null hypothesis, which represents the default assumption (e.g., no difference between groups), and an alternative hypothesis, and then using sample data to determine if there is sufficient evidence to reject the null hypothesis in favor of the alternative.

Confidence Intervals

A confidence interval is a range of values, derived from the sample data, that is believed to cover the true population parameter with a specified probability. In hypothesis testing, if a confidence interval for the difference in means does not include the value of zero (or the null difference), it provides evidence that the means are significantly different. This equivalence underpins the duality of confidence intervals and hypothesis tests.

Transcript

00:01 Hi guys in this problem we have the non -hypothesis h node such that mu -1 equals mu -2 and the alternative hypothesis h1 such that mu -1 does not equal mu -2 okay so we have the n1 equals n2 it's n and 2 n2 equals 15 we have x1 dash equals 4 .7 and x2 i'm not sure it's dash or par it's to x2 equals 7 .8 which is a sample mean and we have variances s1 squared equals 1 and s2 squared equals 6 .25 okay so let's compute the pool standard deviation okay so sp equals the square root of n minus 1 times s 1 squared plus n2 minus 1 times s 2 squared okay over n1 plus n2 minus 2 so after our direct substitution we can get that it's 2 .2638 okay okay under the null hypothesis the test statistics is t where t equals x1 par minus x2 par minus x2 par minus mu 1 minus mu 2, okay, over sp times the square root of 1 over n1 plus 1 over n2.

01:49 Okay, okay.

01:51 So this is 4 .7 minus 7 .8 over 2 .2638 times 1 over 15.

02:08 Plus 1 over 15.

02:11 Okay, so this is negative 3 .75.

02:18 Okay, okay.

02:20 Now we have decrease of freedom.

02:22 Df equals n1, which is 15, plus n2, which is 15, minus 2.

02:28 So it's 28.

02:30 Okay.

02:31 Now let's find the p value using excel.

02:36 So the p value equal is 0 .00081.

02:47 Approximately we can say that it's 0 .001.

02:51 Okay.

02:52 So the p value is less than the significance level, which is 0 .005, so we can reject the null hypothesis.

03:00 Therefore, it can be concluded that there is significant difference of the two mean groups.

03:06 Now let's use the excel function formula.

03:10 And to calculate is a total critical value so t of alpha over 2 and n1 plus n2 minus 2 which is a degree of freedom so it's t of 0 .05 0 .25 and 28 okay so it's 2 .048 okay now it's x1 par minus x2 par plus or minus t of alpha over 2 and n1 plus n 2 minus 2 times sp times 1 over n1 plus 1 over n 2 okay so this equals to negative 4 .793 and negative 1 .407 07 707 7.

04:16 Therefore, the 90 % confidence in interval for the difference of two population means is somewhere between these two values.

04:26 Okay...

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