Statistical Hypothesis testing with confidence interval 3

Question

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.

Answer 1

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 $\mu_1 - \mu_2.$

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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 $(0.275, 5.411).$

var.test(x1,x2)

        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 
          1.220527