Question

I need answer for these in excel format ... please by using excel formulas for example =NORM.dis() ... as soon as possible please! And thankyou!

Fonmattin Undo Cipboard Font A manulacturer produces ight buibs that have a mean ide of at least 600 hours when the production process is working properly. Based on past experience, the population standard deviation is 50 hours and the ight bulb ide is normally distributed The operations manager selects a random sample df5o light bulbs, and the mean life of light bulbs is recorded 10 la. Suppose the populaton mean life SGOhus what เร the probatity that sarple mean Me wil be 11 between 585 and 605 hours? Keep at least 4 decimal places 2 Probability is 13 14 b. Suppose the population mean lite is 600 hours Based on a random samgle of 50 ighi buitbs there tsa 5%chance that the mean life of 1ght bubs wil be aboe L - has 15 16 17 C. Suppose the sample mean Ite of the light bulbs is found to be 580 hours based on a random 8 sample of 50 light bulbs What are the hypotheses that the manager would use to conduct a 19 hypothesis test to see if the population mean amount is diferent from 600 hours? Ho H. Instruction Question O Type here to search 4. 6 2 WİERTY

Answer 1

Part a)

We have population mean $\mu$ = 600 and population standard deviation $\sigma$ = 50 and n = 50

According to sampling distribution of sample mean ( $\bar{x}$ ) ,   $\bar{x}$ follows approximately normal distribution with mean $\mu$ _$\bar{x}$ = 600 and standard deviation $\sigma _{\bar{x}}$ =   $\frac{\sigma }{\sqrt{n}}$ = 50/ sqrt(50) = 7.0711

We are asked to find P( 585 <= $\bar{x}$ <= 605 ) = P( $\bar{x}$ <= 605 ) - P( $\bar{x}$ <= 585)

= [ =NORM.DIST(605,600,7.0711,TRUE)] - [=NORM.DIST(585,600,7.0711,TRUE)]

= 0.7602 - 0.0169

P( 585 <= $\bar{x}$ <= 605 ) = 0.7433

Part b)

We are asked to find $\bar{x}$ such that area above that $\bar{x}$ is 0.05

We have to use excel function =NORM.INV()

The area above $\bar{x}$ is 0.05 , therefore area less than $\bar{x}$ is 1- 0.05 = 0.95

=NORM.INV(0.95,600,7.0711) = 611.63

Part c)

H0 : $\mu$ = 600

Ha : $\mu$ $\neq$ 600

Part d)

We have to perform one sample z test for mean , since here population standard deviation $\sigma$ is known .

We are given $\bar{x}$ = 590

Test statistic z =   $\frac{\bar{x}-\mu }{\sigma /\sqrt{n}}$ = -1.41

Part e)

We are given $\alpha$ = 0.05 , since this is two tailed test $\alpha$ /2 = 0.025

So =NORM.INV(0.025,0,1) = -1.96

Since this is two tail test there are two critical values -1.96 and 1.96

Critical values : -1.96,1.96

Part f)

Since z statistic -1.41 do not fall in the rejection of H0, We fail to reject H0 at 0.05 level of significance

So There is no significant evidence that population mean of light bulb is different that 600 hours

Part g)

P value =2* P( $\bar{x}$ <= -1.41) = 2*NORM.DIST(-1.41,0,1,TRUE)*

P value = 0.1585

It is the probability that sample mean $\bar{x}$ is fall outside the interval ( -1.41 , 1.41)