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Stats Test 2 - Statistics notes for class test two

Statistics notes for class test two

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Applied Statistics (200033)

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Western Sydney University

Academic year: 2020/2021

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Topic 3A

Question 1 Let X be the number of passengers in a car in rush hour traffic in Melbourne. The random variable X is likely to assume the values 0, 1, 2 and 3, with probabilities given as follows

x P (x)

0 0.

1 0.

2 0.

3 0.

Working out:

x P (x) x. p (x) multiply the first column by the second

x^2. p (x) multiply the first column by the third

0 0 0 0

1 0 0 0.

2 0 0 1.

3 0 0 1.

Total 1 1 3.

Calculate the expected number of people in a car. Answer: 3rd column’s total.

Calculate the standard deviation of the number of passengers in a car.

Question 2 A university graduate is thinking of starting a business in the IT industry; but she is worried that she will not make a profit in her first year. She randomly selects a sample of five new IT business start-ups from the past year. It is known that 70% of all business start-ups in the IT industry report a profit in their first year.

From the question, we know Of all IT business start-ups, 70% report a profit in their first year. She selects five new IT business start-ups from the past year. If we let X = number of IT business start-ups, then X ~ Binomial(n = 5, p = 0). 5 = amount of businesses, 0 = percentage of businesses which reported a profit.

(a) Find the probability that none of the start-ups selected reported a profit in their first year.

P(x = 0) = P(x ≤ 0) = 0.

NONE = 0 in the K column.

(b) What is the probability that at least three of the start-ups selected reported a profit in their first year?

P(x ≥ 3) = 1 – P(x 2) = 1 – 0. = 0.

(b) Calculate the probability that in 30 minutes, there will be more than 4 people arriving. X ~ Poisson (μ 12 per hour) X ~ Poisson(μ 6 per 30 minutes)

P(x > 4 | μ 6) = P(x ≥ 5 | μ 6) = 1 – P(x 4 | μ 6) = 1 – 0. = 0.

More than four: 4 in K column 30 Minutes: 30 minutes = half of an hour, the question outlines 12 people arriving per hour, so half of 12 is 6.

(c) The employee wants to have a 10 minute break, but knows that if any one arrives and finds the desk unoccupied, there will be a complaint. If the employee does take a 10 minute break, what is the probability that the shopping mall will receive a complaint? X ~ Poisson(μ 12 per hour) X ~ Poisson(μ 2 per 10 minutes)

P(x > 0 | μ 2) = P(x ≥ 1 | μ 2) = 1 – P(x 0 | μ 2) = 1 – 0. = 0. 1 per 5 minutes as there is 12 per hour. (60 divided by 5)

A complaint means ANY complaint so go to the bare minimum which is 0.

Question 4 Mr. Jones is attempting to determine where he should invest $100,000 that he has won a lottery. He has narrowed his choices down to three investment funds a1, a2, and a3. The events of interest are the various states of the economy. You have constructed the following payoff table showing the various alternatives, states of nature and payoffs.

a1 a2 a

S1 (Recession) $1,000 -$2,000 -$10,

S2 (Stable economy) $4,000 $7,000 $8,

S3 (Economy improves)

$8,000 $12,000 $15,

You estimate the probability of the various states of nature as:

State of Nature Recession Stable Economy Economic Improves

Probability 0 0 0.

(a) Calculate the Expected Profit (EMV) for each possible decision. What would you decide on the basis of maximising the expected profit?

a1 a2 a3 Probability

S1 (Recession) $1,000 -$2,000 -$10,000 0.

S2 (Stable economy)

$4,000 $7,000 $8,000 0.

S3 (Economy improves)

$8,000 $12,000 $15,000 0.

EMV(a1) = S1 Probability(S1) + S2 Probability(S2) + S3 Probability(S3) = TOTAL

EMV(a1) = 0(1000) + 0(4000) + 0(8000) = $4, EMV(a2) = 0(-2000) + 0(7000) + 0(12000) = $6, EMV(a3) = 0(-10000) + 0(8000) + 0(15000) = $6,

EMV* = max(EMV) = $6,700 ← highest number from totals ^

Topic 3B Question 1 Daily balances on personal cheque accounts at the CZ bank are normally distributed, with a mean of $600 and a standard deviation of $150.

(a) What percentage of daily balances are in excess of $900?

Let X be the daily balances; then X ~ N(μ = 600, standard dev. = 150).

> greater than < less than

X is being known as Z for this formula.

(b) What percentage of daily balances are below $320?

X ~ N(μ = 600, standard dev. = 150).

(c) The bank is considering only paying interest to those customers who are carrying daily balances in excess of a certain amount. If it does not want to pay interest to more than 1% of its customers, what is the minimum daily balance on which it should be willing to pay interest? P(x > x0) = 0. P(x < x0) = 0. P(Z < z0) = 0 ← find this on the table

Hence z0 = 2.

x0 = μ + z0 = 600 + 2 150 = 949.

The minimum daily balance on which it should be willing to pay interest is $949.

Question 2 To obtain information on the volume of freight shipped by truck over a particular interstate highway, a state highway department monitored the highway for 25 one hour periods randomly selected throughout a one‐month period. The number of truck trailers was counted for each one‐hour period, and the average was calculated for the sample of 25 individual one‐hour periods. Suppose the number of heavy‐duty trailers per hour is approximately normally distributed, with mean μ = 40 and standard deviation σ = 10

(a) What is the probability that the sample mean for n = 25 one‐hour periods is larger than 45?

X ~ N(μ = 40, = 10)

Then X ~ Bin(n = 144, p = 0). We want to find P(p̂ ≥0). Since np = 5 and nq = 138 are both greater than 5, we can, by the Central Limit Theorem, approximate the sampling distribution of the sample proportion to be normal.

Topic 4A A university lecturer recommends that each student spends at least 20 minuteseach week to review the lecture material in preparation for the class tests.

The lecturer is interested in the average time spent by each student. Hence theytake a sample of 15 students and ask them how long did they spend preparing. The times (in minutes) of the 15 students are as follows:

22, 15, 1, 14, 0, 9, 17, 31, 18, 26, 23, 15, 33, 28, 20

Assuming the times are normally distributed, construct a 95% confidence interval for the population mean time students spend preparing

Question 1 The question asks “construct a 95% confidence interval for the population mean time students spend preparing”

There are two confidence intervals for the population mean depending on whether weknow the population standard deviation or not. Looking at the flow chart in the formulaguide

Is the population standard deviation known? NO. then use...

We want to construct a 95% confidence interval for the population mean timestudents spend preparing. Since is unknown use:

construct a 95% confidence interval: 100 (1 – α) = 95, so ... They take a sample of 15 students ... So (22, 15, 1, 14, 0, 9, 17, 31, 18, 26, 23, 15, 33, 28, 20

So what is the critical value tα/. The degrees of freedom = n – 1. Hence in this question, df = 15 – 1 = 14. So we need to find t 0.

a=0. n= 15 x̅= 18. S= 9.

We want to estimate of the population proportion of retail stores which sell food. If our sample is large

From our question we know With 95% confidence; so α = 0 za/2 = z0.

In a random sample of 1000; so n = 1000

Given 210 sold food; so x = 210

z0 = ??? P(z > z0) = 0. P(z < z0) = 0.

Hence z0 = 1.

The 95% confidence interval for the true proportion of retail stores that sell food is between 0 (18%) and 0 (23%).

Question 3 A national survey research firm has past data that indicate that theinterview time for a consumer opinion study has a standarddeviation of 10 minutes.

How large a sample should be taken if the firm wants to estimate themean interview time to within 1 minutes with 95% confidence?

What do we know? With 95% confidence; so α = 0 z0 = 1. ... to within 1 minutes ... so B = 1. ... has a standard deviation of 10 minutes σ = 10 Estimate μ...

Hence a sample size of 171 is needed.

Question 4 A new cheese product is to be test marketed by giving a free sample torandomly selected customers and asking them to state whether or not theylike the product.

With a 90% confidence level and a target sampling error of 3% or less, what sample size would you recommend if preliminary estimates indicate that approximately 45% of the individuals in the population will like the product?

What do we know? With 90% confidence; so α = 0 z0 = 1.

z0 = ??? P(z > z0) = 0. P(z < z0) = 0.

Hence z0 = (1 + 1) /2 = 1.

What do we know? With 90% confidence; so α = 0 z0 = 1. ... to within 3%... so B = 0.

Estimate a population proportion...

Hence a sample size of 745 is needed.

It is known that selling times are normally distributed, with a populationstandard deviation of 20 days.

A competitor believes that the claim is not true; indeed he believes that the mean selling time of a residential home is greater than 110 days. Heinvestigates by taking a sample of 36 recently sold residential homes, anddetermines the mean selling time of this sample to be 118 days.

Determine the p-value of this test.

Since HA is >, and z = 2 from Question 1... The p-value is P(z > 2) = 1 – P(z < 2) = 1 – 0. = 0 (0%)

NOTE: Since p-value < alpha (0 < 0), we can reject H.

Question 3 A bath soap manufacturing process is designed with the expectation that each batch prepared in the mixing department will produce a mean of 100bars of soap per batch. Quantities over or under this standard areundesirable. A sample of 10 batches shows the following numbers of barsof soap:

88 98 100 102 99 93 104 102 100 103

Assuming normality of soap numbers, test at the 10% significance level tosee if we can conclude that the mean number of bars of soap per batch is not 100.

Question 3

H0: μ = 100 HA: μ ≠ 100
Since σ is unknown:
Here α = 0; and since the test is two tailed and df = n – 1 = 10 – 1 = 9, t0, 9df = 1.
Reject H0 if t > 1 or if t < -1.
𝑥̅= 98 s = 4 n = 10
Since -0 falls between -1 and 1, we cannot reject H0. We cannot conclude that the mean number of bars of soap per batch is not 100

Note: t = -0, hence |t| = 0. Since 0 < 1, we cannot reject H.

0.

Question 4 A car manufacturer conducted a study by randomly sampling andinterviewing 500 consumers in a new target market. The goal of the studywas to determine if consumers would consider purchasing this brand of car. Management has already determined that the company will enter this market. However, if brand preference is lower than 20%, then additional resourceswill be committed to advertising and sponsorship in an effort to enhancebrand awareness amongst the target consumers. In the sample, 85consumers exhibited what the company considered strong brand liking. At the 5% level of significance, can you conclude that additional resourceswill be committed to advertising and sponsorship?

H0 : p = 0. HA : p < 0.
Since n is large:
Here α = 0; and since the test is one tailed, z0 = 1.

... Can you conclude that additional resources will be committed ... ... If brand preference is lower than20%, then additional resources will be committed ... np = 500 × 0 = 100 > 5 nq = 500 × 0 = 400 > 5

Reject H0 if z < -1.
Since -1 < -1, we can reject H. We can conclude that additional resources will be committed to advertising andsponsorship.

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