Question

A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 56 men and 80 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained.Two sample T for Men vs WomenN Mean StDev SE MeanMen 56 112.5 11.1 1.5Women 80 118.7 14.2 1.695% CI for men – mu WomenT-Test mu Men = wu Women (vs<)T = 2.85 P = 0.0025 DF = 132State the null and alternative hypotheses. Which of the following is correct?a)A. H0: u1 = u2; Ha :u1>u2B. H0:u1 = u2;Ha::u1 not equal u2C. H0:u1 = u2;Ha:u1

Answer 1

Answer:

Null hypothesis:$\mu_{m} \geq \mu_{w}$

Alternative hypothesis:$\mu_{m} < \mu_{w}$

$p_v =P(t_{134}$

D. Reject H0, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women.

So on this case the 95% confidence interval would be given by $-10.502 \leq \mu_{m} -\mu_w \leq -1.898$  

We are 95% confident that the mean difference is in the confidence interval.

Step-by-step explanation:

1) Data given and notation

$\bar X_{m}=112.5$ represent the mean for the sample of men

$\bar X_{w}=118.7$ represent the mean for the sample women

$s_{m}=11.1$ represent the sample standard deviation for the sample men

$s_{w}=14.2$ represent the sample standard deviation for the sample eomen

$n_{m}=56$ sample size for the group men

$n_{w}=80$ sample size for the group women

z would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

Null hypothesis:$\mu_{m} \geq \mu_{w}$

Alternative hypothesis:$\mu_{m} < \mu_{w}$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{m}-\bar X_{w}}{\sqrt{\frac{s^2_{m}}{n_{m}}+\frac{s^2_{w}}{n_{w}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

With the info given we can replace in formula (1) like this:

$t=\frac{112.5-118.7}{\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}}=-2.85$  

4) Statistical decision

The degrees of freedom are given by:

$df=n_m +n_w -2= 56+80-2=134$

Since is a left tailed test the p value would be:

$p_v =P(t_{134}$

D. Reject H0, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women.

5) Confidence interval

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_m -\bar X_w) \pm t_{\alpha/2}\sqrt{(\frac{s^2_m}{n_m}+\frac{s^2_w}{n_w})}$ (1)  

The point of estimate for $\mu_1 -\mu_2$ is just given by:  

$\bar X_m -\bar X_w =112.5-118.7=-6.2$  

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,134)".And we see that $z_{\alpha/2}=1.98$  

Now we have everything in order to replace into formula (1):  

$-6.2-1.98\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}=-10.502$  

$-6.2+1.98\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}=-1.898$  

So on this case the 95% confidence interval would be given by $-10.502 \leq \mu_{m} -\mu_w \leq -1.898$  

We are 95% confident that the mean difference is in the confidence interval.