d. In an effort to reduce costs, the assembly instructions were rewritten with fewer pages. The company selects an SRS of 150 purchasers who received the new instructions. What is the probability that at least 25% of the customers in the sample call the help line? normal e. The company finds that 25% of the customers in the sample called the help line. Based on your answer to part (d), do these data provide convincing evidence that the proportion of customers who would call the help line when given the new assembly instructions has increased from 0.18? Explain your reasoning.

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Need help with D and E please
2. A company that sells thousands of bicycles online maintains a telephone help line to assist customers
who are assembling their bicycle. The company keeps detailed records of the percentage of customers
who purchase a bicycle who call in for help. The company claims that 18% of all customers who
purchase a bicycle call the help line.
a. Let the proportion of customers in a sample who call the help line. Calculate the mean and
standard deviation of the sampling distribution of p for random samples of size 150 from this
population.
SD = 0.314
n=150
p=18%
mean = 0.18
SD =
= √(p *g/n)
=
9-1-0.18-0.82
b. Interpret the standard deviation from part (a).
On average, there is a 0.0314 difference between
Sample proportion and the mean
(0,18 0,82/150) = 0.314
c. Would it be appropriate to use a normal distribution to model the sampling distribution of p?
Justify your answer.
np=150*0.18=27
ng=180° 0.82=123
np = 27>10
d. In an effort to reduce costs, the assembly instructions were rewritten with fewer pages. The
company selects an SRS of 150 purchasers who received the new instructions. What is the
probability that at least 25% of the customers in the sample call the help line?
ng=123710 Sampling distribution follows
normal
e. The company finds that 25% of the customers in the sample called the help line. Based on your
answer to part (d), do these data provide convincing evidence that the proportion of customers
who would call the help line when given the new assembly instructions has increased from 0.18?
Explain your reasoning.
Transcribed Image Text:2. A company that sells thousands of bicycles online maintains a telephone help line to assist customers who are assembling their bicycle. The company keeps detailed records of the percentage of customers who purchase a bicycle who call in for help. The company claims that 18% of all customers who purchase a bicycle call the help line. a. Let the proportion of customers in a sample who call the help line. Calculate the mean and standard deviation of the sampling distribution of p for random samples of size 150 from this population. SD = 0.314 n=150 p=18% mean = 0.18 SD = = √(p *g/n) = 9-1-0.18-0.82 b. Interpret the standard deviation from part (a). On average, there is a 0.0314 difference between Sample proportion and the mean (0,18 0,82/150) = 0.314 c. Would it be appropriate to use a normal distribution to model the sampling distribution of p? Justify your answer. np=150*0.18=27 ng=180° 0.82=123 np = 27>10 d. In an effort to reduce costs, the assembly instructions were rewritten with fewer pages. The company selects an SRS of 150 purchasers who received the new instructions. What is the probability that at least 25% of the customers in the sample call the help line? ng=123710 Sampling distribution follows normal e. The company finds that 25% of the customers in the sample called the help line. Based on your answer to part (d), do these data provide convincing evidence that the proportion of customers who would call the help line when given the new assembly instructions has increased from 0.18? Explain your reasoning.
Expert Solution
Step 1

Let p be the population proportion of all customers who purchase a bicycle called the helpline.

Given that,

p = 18% = 0.18

n = 150

To find the probability that the sample proportion is at least 25%.

Test the hypothesis using part d.

 

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