CALCU-A store claims that their average time to pr

A store claims that their average time to process a click and collect order is 90 minutes with a known population standard deviation of 15 minutes. To check their claim, the store randomly sampled 10 customers and found the following processing times: 80, 104, 86, 122, 100, 81, 73, 77, 60, 94 Perform a hypothesis test to check whether the mean processing time, μ, is 90 minutes by completing the following steps. (a) State the null and alternative hypotheses. (b) One of the assumptions of hypothesis testing is that the data comes from a normal distribution. Create a normal probability plot using Matlab. Based on your plot does the assumption of normality seem reasonable? (c) What is our estimate for μ? (d) Calculate the standard error. (e) Calculate the 70% confidence interval for μ. (f) Calculate the value of the test statistic. (g) Calculate the p-value. (h) Do you retain or reject the null hypothesis at the 30% level of significance? Provide justification for your answer. (i) What is your conclusion in the context of the problem? (j) Suppose that you have loaded your data into Matlab with variable name x. What is the command you would use in Matlab to perform the above calculations? (just give the command, not the output)

Solution by Steps

step 1

The null hypothesis (

H_0

) states that the mean processing time is equal to 90 minutes:

H_0: \mu = 90

. The alternative hypothesis (

H_a

) states that the mean processing time is not equal to 90 minutes:

H_a: \mu \neq 90

step 2

To check the assumption of normality, create a normal probability plot using Matlab. The command would be: `normplot(x)`. Based on the plot, if the points closely follow a straight line, the assumption of normality is reasonable

step 3

The estimate for the mean (

\bar{x}

) is calculated as follows:

\bar{x} = \frac{80 + 104 + 86 + 122 + 100 + 81 + 73 + 77 + 60 + 94}{10} = 90.7

step 4

The standard error (SE) is calculated using the formula:

SE = \frac{\sigma}{\sqrt{n}}

, where

\sigma = 15

and

n = 10

. Thus,

SE = \frac{15}{\sqrt{10}} \approx 4.74

step 5

The 70% confidence interval for

\mu

is calculated using the formula:

CI = \bar{x} \pm z \cdot SE

. For a 70% confidence level,

z \approx 1.04

. Therefore,

CI = 90.7 \pm 1.04 \cdot 4.74 \approx (87.4, 94.0)

step 6

The test statistic (

z

) is calculated using the formula:

z = \frac{\bar{x} - \mu_0}{SE} = \frac{90.7 - 90}{4.74} \approx 0.148

step 7

The p-value can be found using the standard normal distribution. For

z = 0.148

, the p-value is approximately

0.440

step 8

At the 30% level of significance (α = 0.30), we compare the p-value with α. Since 0.440 > 0.30, we retain the null hypothesis

step 9

The conclusion is that there is not enough evidence to reject the store's claim that the average processing time is 90 minutes

step 10

The Matlab command to perform the above calculations would be: `ztest(x, 90, 15, 'Alpha', 0.3)`

Answer

The mean processing time is estimated to be 90.7 minutes, and we retain the null hypothesis.

Key Concept

Hypothesis testing involves comparing sample data against a null hypothesis to determine if there is enough evidence to reject it.

Explanation

In this case, we found that the sample mean did not provide sufficient evidence to reject the claim that the average processing time is 90 minutes.

In a certain exam, the scores of the students are normally distributed, with a mean score of 67% and a standard deviation of 7%. (a) What is the probability that a student will fail the exam (i.e. score less than 50%)? Call your answer p, and give it in terms of Φ, the CDF function of the standard normal distribution. (b) Which of the following Matlab commands should be used to calculate the probability in Part (a)? • normcdf(50, 0, 1) • normcdf(1.376, 67, 7) • normcdf(-2.429, 0, 1) • normcdf(-17, 67, 7)

Solution by Steps

step 2

The probability that a student will fail the exam is given by

p = \Phi(-2.43)

, which represents the cumulative distribution function (CDF) of the standard normal distribution at

z = -2.43

step 3

For part (b), the correct Matlab command to calculate the probability in Part (a) is

\text{normcdf}(50, 67, 7)

, which computes the CDF for a normal distribution with mean 67 and standard deviation 7 at the value 50

[1] Answer

A

Key Concept

Normal Distribution and Z-scores

Explanation

The z-score allows us to standardize scores in a normal distribution to find probabilities using the CDF function.

[2] Answer

A

Solution by Steps

step 1

To find the value of

c

that makes

f(x)

a valid PDF, we need to ensure that the integral of

f(x)

over its range equals 1:

\int_1^2 c(x-1)(3x^2 + 2x) \, dx = 1

step 2

We can simplify the integral:

\int_1^2 (x-1)(3x^2 + 2x) \, dx = \int_1^2 (3x^3 + 2x^2 - 3x^2 - 2x) \, dx = \int_1^2 (3x^3 - x^2 - 2x) \, dx

step 3

Evaluating the integral:

\int (3x^3 - x^2 - 2x) \, dx = \left[ \frac{3x^4}{4} - \frac{x^3}{3} - x^2 \right]_1^2

step 4

Calculating the definite integral:

\left( \frac{3(2^4)}{4} - \frac{(2^3)}{3} - (2^2) \right) - \left( \frac{3(1^4)}{4} - \frac{(1^3)}{3} - (1^2) \right) = \left( 12 - \frac{8}{3} - 4 \right) - \left( \frac{3}{4} - \frac{1}{3} - 1 \right)

step 5

After simplifying, we find that the integral evaluates to

\frac{28}{12} - \left( \frac{9}{12} \right) = \frac{19}{12}

. Thus,

c \cdot \frac{19}{12} = 1

gives

c = \frac{12}{19}

step 6

The cumulative distribution function

F(x)

is found by integrating

f(x)

from 1 to

x

:

F(x) = \int_1^x \frac{12}{19}(t-1)(3t^2 + 2t) \, dt

for

1 \leq x \leq 2

step 7

To find P(1.2 < X \leq 1.4) , we calculate

F(1.4) - F(1.2)

Answer

c = \frac{12}{19}

,

F(x)

is the integral of

f(x)

, and P(1.2 < X \leq 1.4) = F(1.4) - F(1.2)

Key Concept

A probability density function (PDF) must integrate to 1 over its range.

Explanation

We found

c

to ensure the PDF integrates to 1, and we set up the cumulative distribution function and probability calculations accordingly.

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