Question

At a recent meeting, it was decided to go ahead with the introduction of a new product if "interested consumers would be willing, on average, to pay $20.00 for the product." A study was conducted, with 315 random interested consumers indicating that they would pay an average of $18.14 for the product. The standard deviation was $2.98. a. Identify the reference value for testing the mean for all interested consumers. b. Identify the null and research hypotheses for a two-sided test using both words and mathematical symbols. c. Perform a two-sided test at the 5% significance level and describe the result. d. Perform a two-sided test at the 1% significance level and describe the result. e. State the p-value as either p>0.05, p<0.05, p<0.01, or p<0.001

Answer 1

There is not enough evidence to support the claim that interested consumers would be willing, on average, to pay $20.00 for the product.

Explanation:

We are given the following in the question:

Population mean, μ = $20.00

Sample mean,
`\bar{x}` = $18.14

Sample size, n = 315

Population standard deviation, σ = $2.98

a) Reference value

`\mu = 20`

b) First, we design the null and the alternate hypothesis

`H_(0): \mu = 20.00\text{ dollars}\\H_A: \mu \\eq 20.00\text{ dollars}`

We use Two-tailed z test to perform this hypothesis.

Formula:

`z_(stat) = \displaystyle\frac{\bar{x} - \mu}{(\sigma)/(√(n)) }`

Putting all the values, we have

`z_(stat) = \displaystyle(18.14 - 20.00)/((2.98)/(√(315)) ) = -11.078`

c) Calculating the p-value at 5% significance level

P-Value < 0.00001

Thus,

p<0.001

d) Calculating the p-value at 1% level of significance

P-Value < 0.00001

Thus,

p<0.001

Thus, at both 5% and 1% level of significance, the p value is lower than the significance level, we fail to accept the null hypothesis and reject it.

There is not enough evidence to support the claim that interested consumers would be willing, on average, to pay $20.00 for the product.