Suppose the coffee industry claimed that the average U. S. adult drinks 1.7 cups of coffee per day. To test this claim, a random sample of 34 adults was selected, and their average coffee consumption was found to be 1.95 cups per day. Assume the standard deviation of daily coffee consumption per day is 0.5 cups. Using a = 0.10, answer the following questions: a. Is the coffee industryâs claim supported by this sample?
b. Determine the p-value for this test.
c. Verify your results using PHStat.

a) This t-value obtained (2.92) is in the rejection region (t > 1.69), hence, the sample does not support the cofdee industry's claim.

b) p-value for this test = 0.006266

c) The p-value obtained for this test is lesser than the significance level at which the test was performed, hence, we can reject the nuĺl hypothesis and say that there is enough evidence to suggest that the coffee industry's claim isn't true based in results obtained from the sample data.

Step-by-step explanation:

a) Degree of freedom = n - 1 = 34 - 1 = 33

The critical value of t for a significance level of 0.10 and degree of freedom of 33 = 1.69

Since we are testing in both directions whether the the average U.S. adult drinks 1.7 cups of coffee per day using our sample,

The rejection region is t < -1.69 and t > 1.69

So, we compute the t-statistic for this sample data to test the claim.

t = (x - μ)/σₓ

x = sample mean = 1.95 cups of coffee per day

μ₀ = The standard we are comparing against = 1.7 cups of coffee per day

σₓ = standard error = (σ/√n)

σ = standard deviation = 0.5 cups

n = Sample size = 34

σₓ = (0.5/√34) = 0.0857492926 = 0.08575

t = (1.95 - 1.70) ÷ 0.08575

t = 2.9154759464 = 2.92

This t-value obtained is in the rejection region, hence, the sample does not support the cofdee industry's claim.

b) Checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 34 - 1 = 33

Significance level = 0.10

The hypothesis test uses a two-tailed condition because we're testing in both directions.

p-value (for t = 2.92, at 0.10 significance level, df = 33, with a two tailed condition) = 0.006266

c) To use PHStat, the claim that the average U.S. adult drinks 1.7 cups of coffee per day is the null hypothesis.

The alternative hypothesis is that the real number of cups of coffee that the average U.S. adult drinks as obtained from the sample data, is significantly different from the 1.7 in the coffee industry's claim.

The p-value obtained from PHstat = 0.0063

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.10

p-value = 0.0063

0.0063 < 0.10

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & say that there is enough evidence to suggest that the coffee industry's claim isn't true based in results obtained from the sample data.

Hope this Helps!!!

75 tickets have been sold so far.

step-by-step explanation:

in this question it is clearly mentioned that 45 children's tickets have been sold and this number is 60% of the total tickets sold.

now let's assume the total number of tickets sold be x.

then 60% of x = 45

⇒

⇒ x(.6) = 45

by dividing with .6 on both the sides of the equation

⇒ x = 75

therefore 75 tickets have been sold so far.

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!