Question

	1970’s	2000’s
	97	99
	105	131
	109	83
	87	95
	96	81
	102	100
	102	88
	114	113
	118	108
	113	116

construct a 99% confidence interval estimate for the mean of the differences. Does this interval contain zero?

• What is the Confidence Interval Estimate:

• Does the interval indicate a difference or not? Explain your answer.

Answer 1

Here in this question, the main objective is to perform the hypothesis testing in whether the means of the samples are the same or not. Hence we will conduct a two-sample t-test to check the hypothesis against the alternative one which is not equal. The test will be a two-sided test with a confidence level of 0.01. (Hence the confidence interval will be 99%)

The hypothesis is stated below:

• Null Hypothesis: M1 – M2 = 0
• Alternative Hypothesis: M1 – M2 ≠ 0

We shall now explain how the test should be constructed:

Using sample data, find the standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic.

• Standard error. Compute the standard error (SE) of the sampling distribution.

  SE = sqrt[ (s₁²/n₁) + (s₂²/n₂) ]

  where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.
• Degrees of freedom. The degrees of freedom (DF) is:

  DF = (s₁²/n₁ + s₂²/n₂)² / { [ (s₁² / n₁)² / (n₁ - 1) ] + [ (s₂² / n₂)² / (n₂ - 1) ] }

  If DF does not compute to an integer, round it off to the nearest whole number. Some texts suggest that the degrees of freedom can be approximated by the smaller of n₁ - 1 and n₂ - 1; but the above formula gives better results.
• Test statistic. The test statistic is a t statistic (t) defined by the following equation.

  t = [ (x₁ - x₂) - d ] / SE

  where x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
• P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a t statistic, use the t Distribution Calculator to assess the probability associated with the t statistic, having the degrees of freedom computed above. (See sample problems at the end of this lesson for examples of how this is done.)

Here, as the number of samples are same then the n1 and n2 will be replaced by n only.

Getting the confidence interval estimate. Here I have used R for calculating the confidence interval estimates. But before that we will discuss how the confidence interval can be constructed:

Here in this formula, z is the z-score for the 99% confidence, Sp is the pooled variance, here we are considering that the variance is same for the two samples, hence the variance is a single value. n1 and n2, we have discussed earlier.

I am attaching the R code for getting the results of the t-test:

t_test = t.test(data$X, data$Y, alternative = c("two.sided", "less", "greater"), mu = 0,
paired = FALSE, var.equal = FALSE, conf.level = 0.99)
t_test

Here data$X is nothing but the first sample and dfata$Y is the second sample.

the result is given below:

Explaining the results:

• Here we see that the p-value is greater than 0.01, hence the Null hypothesis is true. i.e. the sample means are equal.
• the 99% confidence interval is (-14.33106,20.13106) which contains 0
• sample estimates of the first sample are 104.3 and similarly for the second sample: 101.4
• The interval is not indicating that the two samples will have a difference because the interval is very narrow with comparison of the means.

Thanks!!