Hard to understand hypothesis test problem.

Question

In some university were collected 10 data from females and males. Something similar to

M

F

Verify the hypothesis $\mu_M-\mu_F=360g$ knowing that $\mu_M-\mu_F>360$ with $\alpha=.01$

This was an exam problem (I don't recall the exact statement but it was something very similar to this one, the image that I put here is not the one that was in the exam I mean I don't recall the given data).

My solution

This problem is about hypothesis test (I think).

Data:

We have this data coming from some university. Our study parameter is $\mu_M-\mu_F$. And we have $\alpha=.01$

Supposition.

We don't know the distribution of our data but w.l.o.g. let's suppose it follows a normal distribution. (I think I'm wrong here because the sample is small n=10). We also know the variance of $M$ and $F$ (was asked to calculate in before question)

Hypothesis

$H_A=\mu_M-\mu_F>360$

$H_0=\mu_M-\mu_F\le 360$ (Note that here we must have $\le$ althoug was given in the text $\mu_M-\mu_F=360$)

then the test statistic, the region, value of test statistic, conclusion,p value, etc.

I concluded that we should not deny null hypothesis.

Can someone check if what I did is correct?

Am I very wrong?

Thanks in advance for your time

Answer 1

Here is output from R statistical software for the data you give (please proofread), testing $H_0: \mu_F - \mu_M = 0$ against $H_a: \mu_F - \mu_M > 0.$

I don't see how the number 360 is relevant to the data you have posted. Also, for your data, the alternative $H_a: \mu_F - \mu_M < 0$ is uninteresting because $\bar X_F > \bar X_M,$ so there would be nothing to wonder about.

It makes no difference whether the null hypothesis is stated as $H_0: \mu_F - \mu_M = 0$ or as $H_0: \mu_M - \mu_F \ge 0.$ [However, the null hypothesis must always contain an $=$-sign in some form (as $=, \le,$ or $\ge)$ because the equality specifies the exact null distribution.]

This procedure assumes that data for men and women are from normal populations, with the same unknown variance. The procedure is called a 'pooled 2-sample t test'. With data for females listed first, he argrument alte="greater" indicates we are wondering whether female's scores are higher.

m = c(428, 419, 458, 439, 441, 456, 463, 429, 438, 445)
f = c(462, 448, 435, 465, 429, 472, 453, 459, 427, 468)
summary(m); sd(m)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  419.0   431.2   440.0   441.6   453.2   463.0 
## 14.22205  # sample SD of men
summary(f); sd(f)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  427.0   438.2   456.0   451.8   464.2   472.0 
## 16.44384

t.test(f, m, alte="gr", var.eq=T)

        Two Sample t-test

data:  f and m
t = 1.4836, df = 18, p-value = 0.0776
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
 -1.721819       Inf
sample estimates:
mean of x mean of y 
    451.8     441.6 

Although the sample mean 451.8 for women is greater than the sample mean 441.6 for men, the P-value $0.0776 > 0.05$ indicates that it is not 'significantly' greater at the 5% level. We cannot reject $H_0.$

Unless there is strong prior evidence that the population standard deviations of scores for males and females are equal, most practicing statisticians would do the 'Welch separate-variances' version of the two-sample t test, which makes no assumption about equal variances. The R code for this procedure drops the parameter var.eq=T from the procedure t.test. The key line of output is as follows:

t = 1.4836, df = 17.634, p-value = 0.07778

It is typical of the Welch version of the test to have degrees of freedom df smaller than for the pooled version of the test (where degrees of freedom are $\nu = n_1 + n_2 - 2 = 18).$ Because the two sample standard deviations for your data (14.22 and 16.44) are about the same, there is not much difference here between the pooled and Welch versions of the two-sample t test. The P-value for the Welch test is still above 5% so we cannot reject $H_0$.