Question

1. A light bulb manufacturer claims their light bulbs as for 1,000 hours. You sample 50 light bulbs and you find that the sample average light bulb lifetime duration is 955 hours. Assume that you know the population standard deviation, and that it is 220 hours. That is n 50、 x"= 955, σ = 220 Test the null hypothesis Ho: μ-1,000 against the alternative hypothesis Ha: μ < 1,000 at the 0.05 significance level a. Calculate the critical statistic using the qnorm function. Compare this to the test statistic, found using the sample information. Do you reject, or fail to reject the nul hypothesis? Find the p-value associated with the sample statistic under the null hypothesis. You can do this using the pnorm function. Report the p-value. Is your conclusion consistent with part (a)? b. 2. Consider the scenario presented in question 1. Here we will consider how the p-value will change with the sample size, keeping all the other information the same Create a vector, called "n", of different sample sizes, from 10 to 200 at an increment of 5 You can do this using the seq) command Use this vector of sample sizes, along with the information in question1 and the pnormO function, to create a vector of p-values that is the same length as the vector n Plot the vector of sample sizes on the horizontal axis and the vector of p-values on the vertical axis using the plot function Label the axes. Paste the figure below. Wha What range of sample sizes will give you a statistically significant result, holding the population standard deviation and sample average constant? a. t happens to the p-value as the sample size increases? b.

Use R

Answer 1

Answer:

1)

The nul hypothesis is that u l00o. We begin with computing the test statistic xbar 955 # sample mean siama22O > n50 # hypothesized value # population standard deviation, sqrt(O2) #sample size # test statistic L0 1440 We then compute the critical value at 05 significance level apha-025 zapha- qnorml-apha) > Ζ.alpha # critical value pval pnormz) pval- pval L1 O148 # lower tail p-value From above R output, we conclude a level of significance, alpha- O05 from standard normal table, two tailed z apha/2 -l90 since our test is two-tailed reject Ho, if zo < 490 OR if zo 1.90 b. p-value 0.148

2)

a)

> xbar=955 #sample mean
>
> mu0=1000 #hypothesized value
>
> sigma=220 #population standard deviation
>
> n<-seq(10,200,5) # n values from 10 to 200 by margin of 5
>
> pval=rep(0,length(n)) # define p value as same of length of "n"
>
> for(i in 1:length(n)){
+ z= (xbar-mu0)/(sigma/sqrt(n[i]))
+ pval[i]=pnorm(z)
+ }
>
> plot(n,pval,xlab="sample size",ylab="p-value",main = "P value as n increases")

b)

P value as n increases 12 oooo00ooo0000oo0oooo 髦3 100 150 20N) 50 sample size

Comment :- as sample size increases the P value is decreases.i.e result become significant as sample size increases.

c) n >= 65 will give us the statistically significant result.

in R it can be done as

> g<-data.frame(n,pval)
> subset(g,pval<0.05)
n pval
12 65 0.049563762
13 70 0.043508256
14 75 0.038245808
15 80 0.033661609
16 85 0.029659883
17 90 0.026160171
18 95 0.023094476
19 100 0.020405033
20 105 0.018042550
21 110 0.015964801
22 115 0.014135485
23 120 0.012523302
24 125 0.011101189
25 130 0.009845691
26 135 0.008736431
27 140 0.007755678
28 145 0.006887969
29 150 0.006119801
30 155 0.005439364
31 160 0.004836313
32 165 0.004301577
33 170 0.003827193
34 175 0.003406162
35 180 0.003032326
36 185 0.002700262
37 190 0.002405192
38 195 0.002142902
39 200 0.001909671

3)

assuming it is two-tailed test

mu0=1000 #hypothesized value

sigma=220 #population standard deviation, sqrt(102)

n=50 #sample size

xbar <- seq(920,1000,5)
pval = 2* (1- pnorm(abs((xbar-mu0)/(sigma/sqrt(n)))))
pval

plot(xbar,pval)

4)

Let $\mu$ indicate the true mean of the population

We want to test the following hypotheses

$\begin{align*} &H_0:\mu=1000\leftarrow\text{null hypothesis}\\ &H_a:\mu<1000\leftarrow\text{alternative hypothesis}\\ &\alpha=0.025\leftarrow\text{level of significance to test the hypothesis} \end{align*}$

R code to do this left tail test (all statements starting with # are comments)

xBar<-955 #sample mean
mu0<-1000 #hypothesized value
n<-50 #sample size

#a vector of population standard deviations, from 120 to 250
sigma<-seq(120,250,by=5)

#test statistic
z<- (xBar-mu0)/(sigma/sqrt(n))

#p-values, lower tail values P(Z<-z)
pval<-pnorm(z,lower.tail=TRUE)

#plot
plot(sigma,pval,type="l",xlab="Population standard deviation",ylab="p-value",main="p-value vs population sd")

#set the level of significance
alpha<-0.025

#find the range of population standard deviations with p-val<alpha
sigma[pval<alpha]

#get this plot

a)

p-value vs population sd 120 140 160 180 200220 240 Population standard deviation

We can see that the p-value increases as the population standard deviation increases.

We know that the standard error of mean is calculated using

$\begin{align*} \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \end{align*}$

where $\begin{align*} n \text{ is the sample size and }\sigma\text{ is the population standard deviation} \end{align*}$

We can see that the standard error of mean increases with the population standard deviation for a given n.

Next the value of test statistics is

$\begin{align*} z=\frac{\bar{x}-\mu_{H_0}}{\sigma_{\bar{x}}} \end{align*}$

We can see that for a given sample average, the absolute value of z decreases (or the value of negative of z increase) with the increase in standard error (that is increase in population standard deviation)

This is a left tail test. Hence the p-value = P(Z<-z) will only increase when the absolute value of z decrease

b) Out put from R is below

> #find the range of population standard deviations with p-val alpha > sigma [pval<alpha] 1] 120 125 130 135 140 145 150 155 160

we will get statistically significant results when the p-value is less than the level of significance alpha.

For a level of significance alpha=0.025, we can see that we will get statistically significant results when the population standard deviation is between 120 to 160 (approximate), holding the sample average and sample size constant

5)

We reject the null hypothesis when p value<0.05 ie z<-1.645

ie

vn <-1.645

ie

i. <-1.645 *-+ 1000 /50

ie
T 948.82
is the rejection region interms of $\bar{x}$

b)tru_mu<-seq(800,1000,by=10) will generate numbers from 800 to 1000 in intervals of 10

c) power of the test is probability of rejecting the null hypothesis when actually it not true

P(

< 948. 82/μ 900

$= P\left (Z<\frac{948.82-900}{220/\sqrt{50}} \right )=P(Z<1.57)$

When true $\mu=900$ the command interms of qnorm is

norm(1.57) will give you the required probability

so here the required r code is

z<-{948.82-trumu}/{220/sqrt(50)}
z

power<- pnorm(z) will give you the vector of probabilities corresponding to the alternatives from 800 to 1000 in the interval of 10

d) plot( tru_mu,power, main="power curve", ylab="power of the test",xlab="mu")

e) From the graph when mu ranges from 800 to 920 power is greater than 80%

Note: the r codes are given in italics