I want you to conduct a hypothesis test for a difference of means for cholesterol levels between male and female students. There are 148 females and 164 males in our sample. You can treat this as a large sample problem and use z-values for confidence intervals and hypothesis tests (however, Excel uses a t-value in anything it calculates). The output from Microsoft Excel is given below to help. Your job wl be to find the right numbers in the output to help solve the problem EXCEL DESCRIPTIVE STATISTICS FOR CHOLESTEROL LEVELS OF MALES AND FEMALES Females Males Cholesterol Mean 200.318 196.085 Standard Error 0.881 0.966 Median 201 196 Mode 194 196 Standard Deviation 10.721 12.372 Sample Variance 114.939 153.072 0.493 Kurtosis 0.015 0.109 Skewnes:s 0.086 Range 47 61 Minimum 176 166 Maximum 223 227 32158 29647 Sum Count 164 148 Confidence Level (95.0%) 1.742 1.908 Next we want to do a confidence interval for difference of the two means. The Descriptive Statistics are given above. Put a 95% Confidence Interval around the difference of the female mean male mean. This will make clear the approach and meaning of a confidence interval of the difference of two means compared with the confidence interval for each mean. We can treat this as a large sample and use a z-value (If you use a t-value your answer will be the same within the specified rounding error) We will not assume equal variances. We want a 95% Confidence interval for the difference of the two means, what is the BOE for this Confidence Interval. Use 3 significant decimal places and use the proper rules of rounding.

Respuesta :

Answer:

Consider the following explanation

Step-by-step explanation:

Part 1

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

We are given

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Xbar = 200.318

σ = 10.721

n = 148

Lower limit = Xbar - Z*σ/sqrt(n)

Lower limit = 200.318 – 1.96*10.721/sqrt(148)

Lower limit = 200.318 – 1.7272

Lower limit = 198.591

Part 2

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

We are given

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Xbar = 200.318

σ = 10.721

n = 148

Upper limit = Xbar + Z*σ/sqrt(n)

Upper limit = 200.318 + 1.96*10.721/sqrt(148)

Upper limit = 200.318 + 1.7272

Upper limit = 202.045

Part 3

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

We are given

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Xbar = 196.085

σ = 12.372

n = 164

Lower limit = Xbar - Z*σ/sqrt(n)

Lower limit = 196.085 – 1.96*12.372/sqrt(164)

Lower limit = 196.085 – 1.8935

Lower limit = 194.191

Part 4

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

We are given

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Xbar = 196.085

σ = 12.372

n = 164

Upper limit = Xbar + Z*σ/sqrt(n)

Upper limit = 196.085 + 1.96*12.372/sqrt(164)

Upper limit = 196.085 + 1.8935

Upper limit = 197.979

Part 5

Confidence interval for difference is given as below:

Confidence interval = (X1bar – X2bar) ± Z* sqrt[(σ12 / n1)+(σ22 / n2)]

We are given

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence interval = (X1bar – X2bar) ± Z* sqrt[(σ12 / n1)+(σ22 / n2)]

Confidence interval = (200.318 – 196.085) ± 1.96* sqrt[(10.721^2 / 148)+( 12.372^2 / 164)]

Confidence interval = 4.233 ± 1.96* 1.3077

Confidence interval = 4.233 ± 2.563092

Lower limit = 4.233 - 2.563092 = 1.669908

Upper limit = 4.233 + 2.563092 = 6.796092

Confidence interval = (1.670, 6.796)

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