Question

In: Math

The mean and standard deviation of the lifetimes of 9 randomly selected Duracell batteries are 42...

The mean and standard deviation of the lifetimes of 9 randomly selected Duracell batteries are 42 days and 20 days, respectively. The mean and standard deviation of the lifetimes of 11 randomly selected Eveready batteries are 45 days and 17 days, respectively. Suppose that the lifetime of a Duracell battery has NORM(μ_X,σ^2 ) and the lifetime of an Eveready battery has NORM(μ_Y,σ^2 ).

a) Find an unbiased estimate of σ^2.

b) Calculate an 96% confidence interval for μ_X-μ_Y.

c) Test whether the mean lifetime of all Duracell batteries is less than the mean lifetime of all Eveready batteries. State the null and alternative hypotheses, determine the critical region, make a decision based on α=0.05 and write your conclusion.

Solutions

Expert Solution

for sample 1

n1=9

X1bar=42

sd1=20

Same way for sample 2

n2=11

X2bar=45

sd2=17

b)

For 96% conf level of mean we need to perform the z test

Z critical from the table for 96% conf level is -/+2.014091

Z= [(X1bar-X2bar)-(mu1-mu2)]/sqrt[(sd1^2/n1+sd2^2/n2)]

Solving for mu1-mu2 for two different values of z +/- 2.014091 with the given xbar and sd we get

mu1-mu2 = -19.9372

mu1-mu2= 13.9372

So the 96% conf range is -19.9372 to 13.9372

c) Ho: mu1=mu2

Ha: mu11<mu2 (Hypothesis under test)

So.,

finding t stat we get

t= [(X1bar-X2bar)-(mu1-mu2)]/sqrt[(sd1^2/n1+sd2^2/n2)]

where mu1-mu2=0 (null hypothesis)

So., t = -0.357

While the 95% conf critical t value from t table is -1.734064

looking from t table for the area under the curve as 5% with n1+n2-2 = 18 degrees of freedom

So here t value > t critical

and so we can't reject the Ho

So we can conclude that the mean battery life for everady and Duracell are same only.


Related Solutions

Lifetimes of AAA batteries are approximately normally distributed. A manufacturer wants to estimate the standard deviation...
Lifetimes of AAA batteries are approximately normally distributed. A manufacturer wants to estimate the standard deviation of the lifetime of the AAA batteries it produces. A random sample of 23 AAA batteries produced by this manufacturer lasted a mean of 10.1 hours with a standard deviation of 2.2 hours. Find a 90% confidence interval for the population standard deviation of the lifetimes of AAA batteries produced by the manufacturer. Then complete the table below. Carry your intermediate computations to at...
A sample has a mean of M = 42 and a standard deviation of s =...
A sample has a mean of M = 42 and a standard deviation of s = 3.3, and produces a t statistic of t = 2.04. For a two-tailed hypothesis test that has a critical cutoff T of 1.96, do you reject the null or not? a. none of the answers are correct b. there is no way to tell if you can reject or not c. cannot reject the null hypothesis d. reject the null hypothesis
the lifetimes of its tires follow a normal distribution with mean 48,000 miles and standard deviation...
the lifetimes of its tires follow a normal distribution with mean 48,000 miles and standard deviation 5,000 miles. ·a well-labeled sketch of this normal distribution ·the z-score corresponding to 55,000 miles ·the probability that a randomly selected tire lasts for more than 55,000 miles ·the manufacturer wants to issue a guarantee so that 99% of its tires last for longer than the guaranteed lifetime, what z-score should it use to determine that guaranteed lifetime ·the manufacturer wants to issue a...
Given that x is a normal variable with mean 42 and standard deviation 6.2, find the...
Given that x is a normal variable with mean 42 and standard deviation 6.2, find the following probabilities. (Round your answer to four decimal places.) (a) P(x<= 60) (b) P(x>= 50) (c) P(50 <=x<= 60)
A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation...
A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30. Find a 98% confidence interval for the mean price of all textbooks. *
The ages of a group of 135 randomly selected adult females have a standard deviation of...
The ages of a group of 135 randomly selected adult females have a standard deviation of 17.9 years. Assume that the ages of female statistics students have less variation than ages of females in the general​ population, so let sigmaequals17.9 years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics​ students? Assume that we want 95​% confidence that the sample mean is within​ one-half...
The ages of a group of 141 randomly selected adult females have a standard deviation of...
The ages of a group of 141 randomly selected adult females have a standard deviation of 18.9 years. Assume that the ages of female statistics students have less variation than ages of females in the general​ population, so let σ=18.9years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics​ students? Assume that we want 95​% confidence that the sample mean is within​ one-half year...
assume that IQ score are randomly distributed with a mean of 100 and a standard deviation...
assume that IQ score are randomly distributed with a mean of 100 and a standard deviation of 15 . if 49 are randomly find the probability that their mean IQ score is greater than 98
Mean number of desks produced per week is 42 and population standard deviation is 4.67. the...
Mean number of desks produced per week is 42 and population standard deviation is 4.67. the company has introduced new production methods. A random sample of 12 weeks production indicates 44 desks were produced each week. has the introduction of new production methods increased average number of desks produced each week at .05 significance level. Estimate the 95% confidence interval
The duration (in days) of 14 randomly selected space shuttle flights have a standard deviation of...
The duration (in days) of 14 randomly selected space shuttle flights have a standard deviation of 3.54 days. Construct the 99% confidence interval for the population variance and the population standard deviation.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT