In the past, patrons at a cinema spent an average of $ 2.50 for popcorn with standard deviation of $ 0.90. The amount of these expenditures is normally distributed . Following an intensive public campaign about the negative health effects of eating popcorn by a local medical school, the mean expenditures for a sample of 18 patrons is found to be $ 2.10. At 0.01 level of significance, does this recent experience suggest decline in spending

There's no evidence to show that the recent experience suggests a decline in spending.

Step-by-step explanation:

We are given;

Population mean; μ = $2.5

Population Standard deviation; σ = $0.9

Sample mean; x¯ = $2.1

Sample size; n = 18

Significance level; α = 0.01

Let's define the hypothesis;

Null hypothesis; H0: μ ≥ $2.5

Alternative hypothesis; Ha: μ < $2.5

z-score formula is;

z = (x¯ - μ)/(σ/√n)

z = (2.1 - 2.5)/(0.9/√18)

z = -0.4/0.2121

z = -1.886

Using the p-value from z-score calculator attached and using z = -1.886; α = 0.01; one tail; we have;

P-value = 0.2965

This is more than the significance level. Thus, we will fail to reject the null hypothesis and conclude that there is no evidence to show that the recent experience suggests a decline in spending.

goog may be able to you find that answer

126.66 m²

step-by-step explanation:

the pertinent area formula is a = πr², where r is the radius.

here, a = π(6.35 m)² = 40.32π, or approximately 126.66 m²