Sandra asks…

## AP Statistics Help?

It is believed that the average amount of money spent per U.S. **household** per week on food is about $98 with a standard deviation of $10. A random **sample** of 100 households in a certain affluent community yields a mean weekly food budget of $100. We want to test the hypothesis that the mean weekly food budget for all households in this community is higher than the national average.

a) State the hypotheses.

b) Calculate the test statistic, Z. Explain why can you use Z as your test statistic for this problem.

c) Calculate the P-value for the test.

d) Based on your answer in the previous problem, what action do you take with regard to you hypothesis above.

e) Write a sentence to state your conclusion about the mean weekly food budget.

### richmama answers:

Hypothesis Test for mean:

Assuming you have a large enough sample such that the central limit theorem holds, or you have a sample of any size from a normal population with known population standard deviation, then to test the null hypothesis

H0: μ ≤ Δ or

H0: μ ≥ Δ or

H0: μ = Δ

Find the test statistic z = (xbar – Δ ) / (sx / √ (n))

where xbar is the sample average

sx is the sample standard deviation, if you know the population standard deviation, σ , then replace sx with σ in the equation for the test statistic.

N is the sample size

The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.

H1: μ > Δ; p-value is the area to the right of z

H1: μ α, the significance level then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible.

The hypothesis test in this question is:

H0: μ ≤ 98 vs. H1: μ > 98

The test statistic is:

z = ( 100 – 98 ) / ( 10 / √ ( 100 ))

z = 2

we use a z statistics because of two things. With 100 samples it is plausible we have a large enough sample for the mean to be normally distributed, regardless of the underlying distribution. Second, we know the population standard deviation. The student t is used when we have normal data and have to approximate the standard deviation with the sample standard deviation. The student t distribution accounts for the error introduced by the sample standard deviation as an estimator to the population standard deviation.

The p-value = P( Z > z )

= P( Z > 2 )

= 0.02275013

Since the p-value is less than the significance level of 0.05, a very common testing level, we reject the null hypothesis and conclude the alternate hypothesis μ > 98 it true.

Thomas asks…

## Please help with statistics?

It is believed that the average amount of money spent per U.S **household** per week on food is about $98, with standard deviation $10. A random **sample** of 100 households in a certain affluent community yields a mean weekly food budget of $100. We want to test the hypothesis that the mean weekly food budget for all households in this community is higher than the national average.

1. Describe a Type 1 error in the context of this problem. What is the probability of making a Type 1 error?

2. Describe a Type 2 error in the context of this problem. Give two ways to reduce the probability of a Type 2 error.

Any help is highly appreciated thanks!

### richmama answers:

Set out your null and alternative hypotheses. You will need them when you put the errors into context.

A Type I error is when you reject the null hypothesis, but the null hypothesis is true. It happens at a rate equal to alpha, or 100% – the confidence.

A Type II error is when you fail to reject the null hypothesis, but the alternative hypothesis is true. You could reduce the likelihood by increasing sample size or reducing alpha.

Ken asks…

## Statistics, i need answers to these questions! please asap!?

It is believed that the average amount of money spent per u.s **household** per week on food is about 98$, with standard deviation 10$. A random **sample** of 100 households in a certain affluent community yields a mean weekly food budget of 100$. We want to test the hypothesis that the mean weekly food budget for all households in this community is higher than the national average.

1) Perform a significance test at the .05 significance level. Follow the interface toolbox.

2)Describe a type 1 error in the context of this problem. What it the probablity of making a type 1 error?

3)Describe a type 2 error in the context of this problem. Give one way to reduce the probablity of a type 2 error.

4)Describe a power in the context of this problem. Give one way to increase the probablity of power.

### richmama answers:

Hypothesis Test for mean:

Assuming you have a large enough sample such that the central limit theorem holds, or you have a sample of any size from a normal population with known population standard deviation, then to test the null hypothesis

H0: μ ≤ Δ or

H0: μ ≥ Δ or

H0: μ = Δ

Find the test statistic z = (xbar – Δ ) / (sx / √ (n))

where xbar is the sample average

sx is the sample standard deviation, if you know the population standard deviation, σ , then replace sx with σ in the equation for the test statistic.

N is the sample size

The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.

H1: μ > Δ; p-value is the area to the right of z

H1: μ α, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible.

The hypothesis test in this question is:

H0: μ ≤ 98 vs. H1: μ > 98

The test statistic is:

z = ( 100 – 98 ) / ( 10 / √ ( 100 ))

z = 2

The p-value = P( Z > z )

= P( Z > 2 )

= 0.02275013

Since the p-value is less than the significance level we reject the null hypothesis and conclude the alternate hypothesis μ > 98 is true.

Let α be the significance level of the test

consider the following table

_ _ _ _ _ _ Reject H0 _ _ _ _ Fail to Reject H0

H0 is true _ Type I error _ _ _ _ _ ☺ _ _ _

H0 is false _ _ _ ☺ _ _ _ _ _ _ Type II error _

So, a type I error is rejecting H0 when H0 is true, like sending an innocent person to prison

a type II error is letting a guilty person go free after the trial.

P(Type I Error) ≤ α

P(Type II Error) = β

We generally don’t work with Type II errors and instead talk about Power

Power = 1 – P(Type II Error) = 1 – β

in developing tests we try to maximize the Power and minimize α.

A Type I error in this case is saying that the average amount spent is greater than $98 when it is truly less than 98$.

A Type II error is saying that the average is less than $98 when the true value is greater than $98.

Power is the probability of being correct when the null is false. Increased sample size or higher significance level are ways to increase the power.

Carol asks…

## Costs of living?

I’m trying to do a **sample** budget for when I graduate and get a job, but I’m trying to figure out what type of housing, etc. that I’ll be able to afford. What I want to find out specifically is “typical” costs for things like rent, electricity, property tax, utilities, phone, etc., and what a lot of households spend on things like these and groceries, etc. I wonder if there might be websites out there with ballpark figures on how to estimate these things.

### richmama answers:

Http://www.bankrate.com/brm/movecalc.asp

I think this is the best calculator.

But, just google “cost of living calculator” and you’ll get tons.

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