A grocery store has an average sales of $8000 per day.The store introduced several advertising campaigns in order to increase sales.To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 100 days of sales was selected.It was found that the average was $8200 per day.From past information, it is known that the standard deviation of the population is $1500.The p-value is

A) 1.3333.
B) .9082.
C) .0918.
D) .1333.

Question

Jennifer Rider · Accepted Answer

Final Answer : C, Explanation :   The p-value is calculated using the formula for the z-test since the population standard deviation is known. The formula is   z = x ˉ − μ σ / n z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} z = σ / n ​ x ˉ − μ ​  , where   x ˉ \bar{x} x ˉ   is the sample mean,   μ \mu μ   is the population mean,   σ \sigma σ   is the population standard deviation, and   n n n   is the sample size. Plugging in the given values:   z = 8200 − 8000 1500 / 100 = 200 150 = 1.3333 z = \frac{8200 - 8000}{1500/\sqrt{100}} = \frac{200}{150} = 1.3333 z = 1500/ 100 ​ 8200 − 8000 ​ = 150 200 ​ = 1.3333  . The p-value associated with a z-score of 1.3333 (two-tailed test) is approximately 0.0918, indicating the probability of observing a sample mean at least as extreme as $8200, assuming the true population mean is $8000.

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