Inference for Inference for One-Sample One-Sample

Means Means

Steps for doing a confidence Steps for doing a confidence interval:interval:1) State the parameter2) Conditions • 1) The sample should be chosen randomly• 2) The sample distribution should be approximately

normal• - the population is known to be normal, or• - the sample size is large (n 30) (CLT), or• - graph data to show approximately normal• 3) 10% rule – The sample should be less than 10% of the

population• 4) σ is known (or unknown)3) Calculate the interval• If σ is known we perform a z-interval• If σ is unknown we perform a t-interval4) Write a statement about the interval in the

context of the problem.

Confidence interval for a Confidence interval for a population mean (z-interval):population mean (z-interval):

n

zx

*

estimate

Critical value

Standard deviation of the statistic

Margin of error

Formula for a t-confidence interval:Formula for a t-confidence interval:

*s

x tn

estimate

Critical value Standard deviation of statistic

Margin of errorMargin of error

1df n

Degrees of freedom

In a randomized comparative experiment on the effects of calcium on blood pressure, researchers divided 54 healthy, white males at random into two groups, takes calcium or placebo. The paper reports a mean seated systolic blood pressure of 114.9 with standard deviation of 9.3 for the placebo group. Find a 95% confidence interval for the true mean systolic blood pressure of the placebo group.

μ = the true mean systolic blood pressure of healthy white males

State the parameters

Justify the confidence interval needed (state assumptions)

3) The sample should be less than 10% of the population. The population should be at least 540 healthy white males, which I will assume.

2) The sample distribution should be approximately normal. Since n = 54 >30, by the CLT we can assume the sample distribution is approximately normal.

Since the conditions are satisfied a t – interval for means is appropriate.

1) The sample must be random which is stated in the problem.

4) is unknown

We are 95% confident that the true mean systolic blood pressure for healthy white males is between 112.36 and 117.44.

Calculate the confidence interval.114.9x 54n

95% CI

Explain the interval in the context of the problem.

9.3s

53df

sx t

n

9.3114.9 2.009

54

112.36,117.44

Steps for a hypothesis test :Steps for a hypothesis test :

1) Define the parameter2) Hypothesis statements3) Assumptions4) Calculations (Find the p-

value)5) Decision and Conclusion,

1) The sample should be chosen randomly2) The sample distribution should be approximately

normal - the population is known to be normal, or - the sample size is large (n 30) (CLT), or - graph data to show approximately normal (normal

probability plot and box plot)3) 10% rule – The sample should be less than 10% of the

population4) σ is known or unknown

If σ is known we perform a z-testIf σ is unknown we perform t-test

Conditions for one-sample means

Formulas:known:

statistic ofdeviation standard

parameter - statisticstatistictest

z =x

n

x

Formulas:

unknown:

1df n

statistic ofdeviation standard

parameter - statisticstatistictest

t =xs

n

x

Example 2 The Fritzi Cheese Company buys milk from several suppliers as the essential raw material for its cheese. Fritzi suspects that some producers are adding water to their milk to increase their profits. Excess water can be detected by determining the freezing point of milk. The freezing temperature of natural milk varies normally, with a mean of -0.545 degrees and a standard deviation of 0.008. Added water raises the freezing temperature toward 0 degrees, the freezing point of water (in Celsius). The laboratory manager measures the freezing temperature of five randomly selected lots of milk from one producer with a mean of -0.538 degrees. Is there sufficient evidence to suggest that this producer is adding water to his milk?

Parameters and Hypotheses

μ = the true mean freezing temperature of milk

Assumptions (Conditions)

Since the conditions are met, a z-test for the one-sample means is appropriate.

1) The sample must be random which is stated in the problem.

H0: μ = -0.545

Ha: μ > -0.545

3) The sample should be less than 10% of the population. The population should be at least 50 lots of milk, which I will assume.

2) The sample distribution should be approximately normal. Since it is stated in the problem that the population is normal then the sample distribution is normal.

4) is known

( 1.9565) .0252p value P z

Calculations

.0252 .05

.538x 5n

= 0.05

.008

.545x x

xz

n

.538 ( .545).008

5

1.9565

Conclusion:

Decision: Since p-value < , I reject the null hypothesis at the .05 level.

There is sufficient evidence to suggest that the true mean freezing temperature is greater than -0.545. This suggests that the producer is adding water to the milk.

Example 3 (page 545 #33)

In 1998, as an advertising campaign, the Nabisco Company announced a “1000 Chips Challenge,” claiming that every 19-ounce bag of their Chips Ahoy cookies contained at least 1000 chocolate chips. Dedicated Statistics students at the Air Force Academy (no kidding) purchased some randomly selected bags of cookies, and counted the chocolate chips. Some of their data are give below.

What does this say about Nabisco’s claim. Test an appropriate hypothesis.

1219 1214 1087 1200 1419 1121 1325 1345

1244 1258 1356 1132 1191 1270 1295 1135

Parameters and Hypotheses

μ = the true mean number of chocolate chips in each bag of Chips Ahoy

H0: μ = 1000Ha: μ > 1000

Assumptions (Conditions)

Since the conditions are met, a t-test for the one-sample means is appropriate.

1) The sample must be random which is stated in the problem.

3) The sample should be less than 10% of the population. The population should be at least 160 bags of Chips Ahoy, which we will assume.

2) The sample distribution should be approximately normal.

4) is unknown

The normal probability plot is fairly linear and the boxplot shows no outliers, so we will assume that the sample distribution is approximately normal.

8( 10.1053) 2.176 10p value P t

Calculations

82.176 10 .05

x

16n

= 0.05

s

1000x x

xt

s

n

(1000)

16

10.1053

15df

1238.1875

94.282

1238.1875

94.282

Conclusion:

Decision: Since p-value < , I reject the null hypothesis at the .05 level.

There is sufficient evidence to suggest that the true mean number of chocolate chips in each bag of Chips Ahoy is greater than 1000.