We substitute the given values into the t -statistic formula: sample mean ( x) = 147, hypothesized population mean () = 140, sample standard deviation (s) = 16, and sample size (n) = 26. After performing the calculations, we find the t -statistic to be approximately 2.231.

We substitute the given values into the t -statistic formula: sample mean ( x) = 147, hypothesized population mean () = 140, sample standard deviation (s) = 16, and sample size (n) = 26. After performing the calculations, we find the t -statistic to be approximately 2.231.

A soap manufacturing company was distributing a particular brand of a soap through a large number of retail shops. Before a heavy advertisement campaign, the mean sales per week per shop was 140 dozen. After the campaign, a sample of 26 shops was taken and mean sales was found to be 147 dozen with standard deviation 16. Can you consider the advertisement campaign effective ? [Given $t_{25}(0 \cdot 05) = 2 \cdot 06$ ]

Answer: Yes, the advertisement campaign can be considered effective.

Step-by-step solution

Step 1: Formulate Hypotheses

We set up the null hypothesis ( $H_0$ ) that the advertisement campaign had no effect, meaning the mean sales remain 140 dozen. The alternative hypothesis ( $H_1$ ) is that the campaign was effective, leading to an increase in mean sales, so $\mu > 140$ . This is a one-tailed test.

Step 2: Calculate the Test Statistic

We use the $t$ -distribution since the sample size is small ( $n < 30$ ) and the population standard deviation is unknown. The formula for the $t$ -statistic is given by the sample mean minus the hypothesized population mean, divided by the sample standard deviation over the square root of the sample size.

Step 3: Substitute Values and Compute t-statistic

We substitute the given values into the $t$ -statistic formula: sample mean ( $\bar{x}$ ) = 147, hypothesized population mean ( $\mu$ ) = 140, sample standard deviation ( $s$ ) = 16, and sample size ( $n$ ) = 26. After performing the calculations, we find the $t$ -statistic to be approximately 2.231.

Step 4: Compare with Critical Value

The degrees of freedom for this test are $n-1 = 26-1 = 25$ . For a one-tailed test with a significance level of 0.05 and 25 degrees of freedom, the critical $t$ -value is given as 2.06. We compare our calculated $t$ -statistic with this critical value.

Step 5: Make a Decision

Since our calculated $t$ -statistic (2.231) is greater than the critical $t$ -value (2.06), we reject the null hypothesis. This means there is sufficient evidence to conclude that the advertisement campaign was effective.

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