Question

1. The number of hot dogs sold per game at one concession stand at a baseball park is normally distributed with μ = 23,750 and σ = 108. You select 16 games at random, and the mean number of hot dogs sold, top enclose x, is denoted. Find the probability that the mean number of hot dogs sold for a random sample of 16 games will be less than 23,700 hotdogs. Round your answers to four decimal places.

2. The number of hours spent per week on household chores by all adults has a mean of 26.2 hours and a standard deviation of 5.0 hours. The probability, rounded to four decimal places, that the mean number of hours spent per week on household chores by a sample of 100 adults will be more than 26.75 is: (Do not round your intermediate calculations.)

3. A company that manufactures light bulbs claims that its light bulbs last an average of 1,150 hours. A sample of 25 light bulbs manufactured by this company gave a mean life of 1,097 hours and a standard deviation of 133 hours. A consumer group wants to test the hypothesis that the mean life bulbs produced by this company is less than 1,150 hours. Assume the population is normally distributed.

The test statistic for this hypothesis test is -1.99. What is the p-value for this hypothesis test, rounded to the nearest ten thousandth (e.g., 0.1234)?

4. More and more people are abandoning national brand products and buying store brand products to save money. The president of a company that produces national brand coffee claims that 40% of the people prefer to buy national brand coffee. A random sample of 700 people who buy coffee showed that 259 of them buy national brand coffee. You would like to test whether the percentage of people who buy national brand coffee is different from 40%. Calculate the test statistic for this hypothesis test. Round your answer to the nearest hundredth (e.g., 1.23).

Answer #1

1. µ = 23750, σ = 108, n = 16

P(X̅ < 23700) =

= P( (X̅-μ)/(σ/√n) < (23700-23750)/(108/√16) )

= P(z < -1.8519)

Using excel function:

= NORM.S.DIST(-1.8519, 1)

= **0.0320**

----

2. µ = 26.2, σ = 5, n = 100

P(X̅ > 26.75) =

= P( (X̅-μ)/(σ/√n) > (26.75-26.2)/(5/√100) )

= P(z > 1.1)

= 1 - P(z < 1.1)

Using excel function:

= 1 - NORM.S.DIST(1.1, 1)

= **0.1357**

----

3. x̅ = 1097, s = 133, n = 25

Null and Alternative hypothesis:

Ho : µ = 1150

H1 : µ < 1150

Test statistic:

t = (x̅- µ)/(s/√n) = (1097 - 1150)/(133/√25) = -1.9925

df = n-1 = 24

p-value = T.DIST(-1.9925, 24, 1) = **0.0289**

----

4. n = 700

x = 259

p̂ = x/n = 0.37

Null and Alternative hypothesis:

Ho : p = 0.4

H1 : p ≠ 0.4

Test statistic:

z = (p̂ -p)/√(p*(1-p)/n) = (0.37 - 0.4)/√(0.4 * 0.6/700) =
**-1.62**

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