Question

The price of a certain combo meal at different franchises of a national fast food company varies from​ $5.00 to ​$17.37and has a known standard deviation of

$2.09. A sample of 29students in an online course that includes students across the country stated that their average price is $6.00.

The students have also stated that they are generally unwilling to pay more than $6.75 for this meal. Formulate and conduct a hypothesis test to determine if you can conclude that the population mean is less than $6.75.

Use a level of significance of 0.05.

Is there sufficient evidence at the 0.05 level of significance that the population mean is less than ​$6.75​?

Determine the null​ hypothesis, H0​,and the alternative​ hypothesis, H1.

compute test statistic

find the pvalue

state the conclusion

Answer 1

A sample of 29 students in an online course that includes students across the country stated that their average price is $6.00.

It has a known standard deviation of $2.09.

Here n = 29    ,    = 6      ,    = 2.09

Since standard deviation is known , so we can use z-score .

The students have also stated that they are generally unwilling to pay more than $6.75 for this meal. Formulate and conduct a hypothesis test to determine if you can conclude that the population mean is less than $6.75.

To test

H0 : 6.75               { population mean is less than equal to $6.75 }

H1 : > 6.75                { population mean is greater than $6.75 }

Test Statistics TS:

TS =

Now = 6.75   ,   n = 29    ,    = 6      ,    = 2.09

Hence

TS   =  

       =

TS = -0.75 / 0.3881033 = -1.932475

So calculated test statistics is TS = -1.932475

Is there sufficient evidence at the 0.05 level of significance that the population mean is less than ​$6.75​?

- Since our alternative hypothesis is of " > " type i.e one tail .

so rejection criteria is given by .

we reject null hypothesis if calculatd test statistics is greater than Z-critical value -

Now for 0.05 level of significance Critical is - = - 1.64

That we we reject null hypothesis if TS   > - 1.64

now TS = -1.932475 = 1.932475

Hence TS = -1.932475 < - 1.64

So   TS <

Hence we fail to reject null hypothesis H0 at 0.05 level of significance .

To compute P-value

Since our alternative hypothesis is of " > " type i.e one tail . then p-value is given by

P-Value = P( Z > TS )

            = P( Z > -1.932475 )

          = 1 - P( Z < -1.932475 )

here Z ~ N(0,1)

It can be obtained from R as follow            

> 1-pnorm(-1.932475)               # 1 - P( Z < -1.932475 )
[1] 0.9733495

So P-Value = 0.97 > 0.05

Hence we do not reject null hypothesis .

Yes , there sufficient evidence at the 0.05 level of significance that the population mean is less than ​$6.75​

State the conclusion

- The mean price of a certain combo meal at different franchises of a national fast food company that is the population mean is less than $6.75.