Hi I´m working of the following problem and I got stuck , I hope you can help me.

The exericse is related with the time of manufacturing in a business, this is the table given here---> Table

There are two samples of size 9 (method 1 and method 2), I have obtained their sample mean and their sample variance $$ \mu_{0}=35.22 \hspace{2cm}\mu_{1}= 31.56$$ $$S_{0}^{2}=21.73 \hspace{2cm} S_{1}^{2}=17.8$$ (both samples are normally distributed).

The exercise asks for: Using a confidence interval, prove if it is possible or not with a $95\%$ confidence assume that both population variances are the same. s With the information obtained before, prove whether or not the following statement is true: Method 2 is better than Method 1, in the sense that it reduces the time of manufacturing.

My attempt: I say that $H_{0}\equiv \sigma_{1}^{2}/\sigma_{2}^{2}=1$ (quotient of population variances) is my hypothesis, then I estimate that the confidence interval is $(0.2754.5.4121)$, so as $1$ is contained in the interval we can assume that both variances are the same.

I dont know how to continue with the second part of the exercise, besides, I´m not quite sure about the last interval.

Thank you for your time.


First, I get the same sample means you do from the linked data, but different sample variances. I will use my variances.

x1 = c(32,37,35,28,41,44,35,31,34)
x2 = c(35,31,29,25,34,40,27,32,31)

mean(x1);  var(x1);  mean(x2);  var(x2)
## 35.22222  # mean and var for Samp 1
## 24.44444
## 31.55556  # mean and var for Samp 2
## 20.02778

Second, I think you are being asked to do a pooled 2-sample t test of the null hypothesis that the population means are equal $H_0: \mu_1 = \mu_2$ against the alternative that Method 2 has significantly smaller manufacturing times than Method 1 $H_a: \mu_1 > \mu_2.$ (The pooled test assumes that the population variances are equal.)

There seems to be a problem with your numbering: 0 and 1 in one place and 1 and 2 in another. In view of $\bar X_1 = 35.22$ and $\bar X_2 = 31.55,$ it makes no sense to test whether data indicate Method 1 has smaller times than Method 2.

In R statistical software, I get the following output for the one-sided pooled t test.

t.test(x1, x2, alte="gr", var.eq=T)

        Two Sample t-test

data:  x1 and x2
t = 1.6495, df = 16, p-value = 0.05927
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
 -0.2142871        Inf
sample estimates:
mean of x mean of y 
 35.22222  31.55556 

Somewhere in your text or notes, I suppose you will find a formula for the pooled t test statistic, here $T = 1.6495.$ You should find the formula and verify the value of the $T$ statistic. In the formula you will need a 'pooled variance estimate'; here (with $n_1 = n_2)$ it amounts to $S_p^2 = (S_1^2 + S_2^2)/2 = 22.24.$

In order to find the critical value for a test at the 5% level, you will need the degrees of freedom (df) $9+9-2 = 16$ as shown in the printout. Looking at row df = 16 of a printed table of t distributions, you will find that value to be about $q^* = 1.746.$ Because your $T = 1.6495 < 1.746,$ you cannot reject $H_0$ at the 5% level. That is to say $\bar X_2$ is smaller then $\bar X_1,$ but not by enough to be statistically significant. The 'true difference' in the printout refers to the difference $\mu_1 - \mu_2$ in population means.

Two other features of the printout above also indicate that you cannot reject $H_0$ at the 5% level:

  • The P-value is greater than 5%, indicating that the difference between the sample means is not 'surprisingly' large.

  • The one-sided 95% confidence interval $(-.214, \infty)$ contains $0,$ indicating that $0$ is a believable value of $\mu_1 - \mu_2.$

Notes: (1) Because there was some confusion in the computations and notation in your Question as posted, I cannot be sure I have answered all the questions you may have in mind. After you have had a chance to digest my answer, maybe you will want to leave a Comment with additional questions, and maybe one of us can answer them.

(2) I'm not sure why, but you tried checking whether the population variances $\sigma_1^2$ and $\sigma_2^2$ are equal. Because your sample variances were incorrect, your interval is also incorrect. The R output for a test of equality of variances is shown below; the correct 2-sided 95% CI for the ratio of variances is shown to be $(0.275, 5.411).$


        F test to compare two variances

data:  x1 and x2
F = 1.2205, num df = 8, denom df = 8,
p-value = 0.7849
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 0.2753114 5.4109136
sample estimates:
ratio of variances 
  • $\begingroup$ Thank your four time and effor @BruceET I really appreciate, I obtained again the variances and the interval, and my solution is identical to yours. I have to apologize for the bad notation used in the question, sorry. The exercise asks to prove or disprove if the poblational mean of the second method is lower than the first method. You have proved that we can not reject $H_{0}$ , it means that the poblational means are the same ? On the other hand, i have to assume that poblational variances are the same in order to apply a two sample t-test. $\endgroup$ – user489150 Dec 20 '17 at 15:06
  • $\begingroup$ We have obtained a $95\%$ confidence interval for the poblational variances, and another $95%$ confidence interval for the poblational means, does it mean that we can assure that the poblational means are equal with a $(95)^{2}\%$ of reliability? Thank you. $\endgroup$ – user489150 Dec 20 '17 at 15:06
  • 1
    $\begingroup$ What is 'problational'? // Failure to reject the null hypothesis that two parameters are equal does not mean they are equal. It means we do not have evidence they are different. $\endgroup$ – BruceET Dec 21 '17 at 1:04

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