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 t25(0⋅05)=2⋅06]
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Step-by-Step Solution
Step 1: Formulate Hypotheses
We set up the null hypothesis (H0) that the advertisement campaign had no effect, meaning the mean sales remain 140 dozen. The alternative hypothesis (H1) is that the campaign was effective, leading to an increase in mean sales, so μ>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 (xˉ) = 147, hypothesized population mean (μ) = 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.