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LUCKY_DIMON [66]
2 years ago
8

Monogram Masters advertises on its website that 92% of customer orders are received within five working days. They performed an

audit from a random sample of 200 of the 3,000 orders that month and it shows 175 orders were received on time.
Part A: Can we use a normal approximation? Explain. (5 points)

Part B: If Monogram Masters customers really receive 92% of its orders within five working days, what is the probability that the proportion in the random sample of 200 orders is the same as the proportion found in the audit sample or less? (5 points) (10 points)
Mathematics
1 answer:
Pavlova-9 [17]2 years ago
5 0

Answer:

yes

0.0094

Step-by-step explanation:

p = 0.92, n = 200

np = 184 > 10

n(1 − p) = 16 > 10

Therefore, normal approximation can be used.

μ = p = 0.92

σ = √(p (1 − p) / n)

σ = √(0.92 × 0.08 / 200)

σ = 0.0192

x = 175/200 = 0.875

z = (x − μ) / σ

z = (0.875 − 0.92) / 0.0192

z = -2.35

P(z < -2.35) = 0.0094

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If carla wants to spend the least amount of money she should go to the Braves game because it has one of the lowwest admission and the food is cheaper than any where else. Yes, the concert admission is the same price but the tee-shirts and soda costs more.

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The results of a mathematics placement exam at two different campuses of Mercy College follow: Campus Sample Size Sample Mean Po
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Answer:

z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37  

p_v =P(Z>3.37)=1-P(Z  

Comparing the p value with a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.  

Step-by-step explanation:

Data given

Campus   Sample size     Mean    Population deviation

   1                 330               33                      8

   2                310                31                       7

\bar X_{1}=33 represent the mean for sample 1  

\bar X_{2}=31 represent the mean for sample 2  

\sigma_{1}=8 represent the population standard deviation for 1  

\sigma_{2}=7 represent the population standard deviation for 2  

n_{1}=330 sample size for the group 1  

n_{2}=310 sample size for the group 2  

\alpha Significance level provided  

z would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for Campus 1 is higher than the mean for Campus 2, the system of hypothesis would be:

Null hypothesis:\mu_{1}-\mu_{2}\leq 0  

Alternative hypothesis:\mu_{1} - \mu_{2}> 0  

We have the population standard deviation's, and the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:  

z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

With the info given we can replace in formula (1) like this:  

z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37  

P value  

Since is a one right tailed test the p value would be:  

p_v =P(Z>3.37)=1-P(Z  

Comparing the p value with a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.  

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The daily consumption of milk in a house is 3.25 litres. How much milk will be consumed in 30 days?
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You are given a rate at which the milk is consumed (3.25 litres per 1 day). This is just multiplied by the 30 days to get your answer.

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