Question

A melting point test of n = 10 samples of a binder used in manufacturing a rocket propellant resulted in a sample mean of 154.3 °F. Assume that the melting point is normally distributed with σ = 1.5°F.

A) Is it possible to conclude at 1% significance that the melting point is not 155°F? What is the value of z_calc ?

B) Find a lower value in a 95% confidence interval for the melting point?  

Answer 1

Solution :

Given that ,

A )

= 155°F

= 154.3 °F.

=  1.5°F.

n = 10

The null and alternative hypothesis is ,

H₀ :   = 155°F

H_a :    155°F

This is the two tailed test

Test statistic = z

= ( - ) / / n

= ( 154.3 - 155 ) / 1.5 / 10

= -1.48

The value of the test statistic Z = -1.48

B )

Given that,

Point estimate = sample mean = = 154.3

Population standard deviation =    = 1.5

Sample size = n = 10

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96}

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 1.5 /  10 )

= 0.93

At a lower value 95% confidence interval estimate of the population mean is,

- E

154.3 - 0.93

( 153.37 )

The lower value 95% confidence interval of the population mean is : ( 153.37 )