Question

For each of the following, (a) state the null hypothesis and (b) state the alternate hypothesis, (c) state the type of test that you will use to test the hypothesis and the level of significance, (d) state the critical value , (e) calculate the z-statistic or the t-statistic, (f) state whether we shall reject or fail to reject the null hypothesis, (g) state the relation between the p-value and α with an appropriate inequality, and (h) conclude the hypothesis test with an appropriate summarizing statement addressing the alternate hypothesis.

1. Rosie is an aging cocker spaniel in Elkin, NC who gets regular check-ups from her owner, the local veterinarian. Let x be a random variable that represents Rosie’s resting heart rate (in beats per minute). The vet checked the Merck Veterinary Manual and found for dogs of this breed, μ = 115 beats per minute. Over the past six weeks, Rosie’s heart rate in beats per minute measured 93, 109, 110, 89, 112, and 117. The vet is concerned that Rosie’s heart rate may be slowing. At the α = 0.01 level of significance, do the data indicate that this is the case? A normal probability plot shows the data to have a linear correlation of 0.938. A box-and-whiskers plot of the data shows no outliers.

2. In 1990, 23.8% of the 18- to 24-year old population of Finy Beach was attending college. This year, a random sample of 610 Finy Beach residents in the age group indicated showed that 178 were attending college. Do these data indicate that the population proportion of the 18- to 2y-year olds in Finy Beach who are attending college is different from 23.8%? Test the claim using α = 0.05.

Answer 1

1) The null hypothesis is Ho : mu >=115

Ha : mu < 115

Level of significance is 5% and test is one sided.

Test statistic is t = n^1/2 *(Xbar - mu) /s with degree of freedom n-1

X bar = sample mean = 105 (calculated)

n = sample size = 6

s = sample standard deviation = 11.26 ( calculated with df =5)

Calculated value of t = -10* 2.449 / 11.26 = -2.17

The mod of this value is 2.17

The pvalue is 0.041072

The Critical value of t is calculated with degree of freedom =5 and alpha =0.05 This is found to be 2.015 (from t table)

As mod calculated t > t tabulated we reject the null hypothesis

Also as p < level of significance (0.041072 <5) we reject the null hypothesis

Hence we conclude that there is a significant slowing in the heart beat of the cocker spaniel.

2) Here Ho : P =0.238

Ha : p is not equal to 0.238

Level of significance is 5% (two sided)

Test statistic is Z = p-P /(PQ/n)^1/2

where p = sample proportion = 178/610 = 0.291

P = population proportion = 0.238

Q= 1-P =0.762

n= 610

Putting this we get Z =3.08

Critical value of Z is 1.96 (corresponding to right tail of 2.5%)

p value of getting a Z score of 3.06 is 0.00207

Hence as Zcalc > Z tab we reject the null hypothesis

Also as p < 2.5% we reject the null hypothesis

We thus conclude that the proportion is significantly different form 23.8%