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.

f. Find the p-value using statistical software

Answer 1

SolutionA:

on average, to pay $20.00 for the product

SolutionB:

interested consumers will be willing, on average, to pay $20.00 for the product

Null hypthesis:

H0:

Alternative hypothesis:

interested consumers will not be willing, on average, to pay $20.00 for the product

Ha:

c. Perform a two-sided test at the 5% significance level and describe the result.

t=xbar-mu/s/sqrt(n)

t=(18.14-20)/2.98/sqrt(315)

=-11.08

test statistic=-11.08

df=n-1=315-1=314

p=0.000

p<0.05

Reject H0

Acccept H1

Conclusion:

there is no suffcient evidence at 5% level of signifcance to support the claim that interested consumers will be willing, on average, to pay $20.00 for the product.

Solutiond:

t=(18.14-20)/2.98/sqrt(315)

=-11.078

test statistic=-11.08

df=n-1=315-1=314

p=0.000

p<0.01

conclusion:

There is no suffcient evidence at 1% level of signifcance to support the claim that interested consumers will be willing, on average, to pay $20.00 for the product.

e. State the p-value as either p > 0.05, p < 0.05, p < 0.01, or p < 0.001.

p<0.05,p<0.01

f. Find the p-value using statistical software

mu <- 20
s <- 2.98
n <- 315
xbar <- 18.14
t <- (xbar-mu)/(s/sqrt(n))
t
pvalue <- 2*pt(-abs(t),df=n-1)
pvalue

output:

1] -11.07776
> pvalue <- 2*pt(-abs(t),df=n-1)
> pvalue
2.703362e-24
>  
>
> 2*pt(-abs(t),df=n-1)
2.703362e-24