Question

PART ONE: A toy manufacturer wants to know how many new toys children buy each year. A sample of 305 children was taken to study their purchasing habits. Construct the 80% confidence interval for the mean number of toys purchased each year if the sample mean was found to be 7.6. Assume that the population standard deviation is 1.5. Round your answers to one decimal place.

PART TWO: Find the area under the standard normal curve between z=−0.75z=−0.75 and z=1.83z=1.83. Round your answer to four decimal places, if necessary.

PART THREE: The mean points obtained in an aptitude examination is 188 points with a variance of 361.

What is the probability that the mean of the sample would be less than 186.6 points if 73 exams are sampled? Round your answer to four decimal places.

PART FOUR:

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 444 gram setting. It is believed that the machine is underfilling the bags. A 43 bag sample had a mean of 436 grams. Assume the population standard deviation is known to be 21. A level of significance of 0.05 will be used. State the null and alternative hypotheses.

Answer 1

1)

mean = 7.6 , s = 1.5 , n = 305

z value at 80% confidecne interval is,

alpha = 1 - 0.80 = 0.20
alpah/2 = 0.20/2 = 0.10
Zalpha/2 = Z0.10 = 1.2816

MArgin of error = E = z *(s/sqrt(n))
= 1.282 * (1.5/sqrt(305))
= 0.1101

The 80% confidenc einterval is

mean - E < mu <mean +E

7.6 - 0.1101 < mu < 7.6+ 0.1101
7.5 < mu <7.7

2)

P(-0.75< z < 0.75)
= P(z< 0.75 ) - P(z < -0.75)
= 0.7734 - 0.2266
= 0.5468

P(-1.83 < z < 1.83)
= P(z< 1.83) - P(z < -1.83)
= 0.9664 - 0.0336
= 0.9328

3)

mena = 188 , sigma = sqrt(361) = 19

n = 73

P(x < 186.6)
= P(z < (x - mean)/(sigma/sqrt(n))
= P(z < (186.6 - 188)/(19/sqrt(73))
= P(z < -0.6296)
= 0.2645