Please correct for ford Taurus:?

Question

Let X be a random variable that represents assembly time for the Ford Taurus. The Wall Street Journal reported that the average assembly time is µ = 38 hours. A modification to the assembly proedure has been made. Experience with this new method indicates that σ =1.2 hours. It is thought that the average assembly time may be reduced by this modification. A random sample of 47 new Ford Taurus automobiles coming off the assembly line showed the average assembly time using the method to be x =37.5 hours. Does this indicate that the average assembly time has been reduced? Use σ = 0.01. please make corrections: Data given:

n = 47, y = 38, s = 1.2, u0 = 37.5, a = 0.1;

Step 1 - State the Hypotheses:

Ho: u = 37.5

Ha: u > 37.5

Step 2 - State the Test Statistic:

t = (y-u0)/(s/root(n))

Step 3 - State the Critical or Rejection Region:

The critical region depends upon Ha.

For t > 37.5, we reject Ho if

t > tn-1,a

t > t(46,0.1)

t > 0

Step 4 - Conduct Experiment and Calculate Test Statistic:

t = (y-u0)/(s/root(n))

t = (38-37.5)/(1.2/root(47))

t = 2.857

Step 5 - Reach Conclusions and State in English:

Since t > 0, we have sufficient evidence to reject Ho. We therefore have enough evidence to suggest that the true mean is more than 37.5.

DONE.

Answer 1

your hypothesis is not correct, the hypothesized values are 38, not 37.5.  the 37.5 is the sample data.

Hypothesis Test for mean:

Assuming you have a large enough sample such that the central limit theorem holds, or you have a sample of any size from a normal population with known population standard deviation, then to test the null hypothesis

H0: μ ≤ Δ or

H0: μ ≥ Δ or

H0: μ = Δ

Find the test statistic z = (xbar - Δ ) / (sx / √ (n))

where xbar is the sample average

sx is the sample standard deviation, if you know the population standard deviation, σ , then replace sx with σ in the equation for the test statistic.

n is the sample size

The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.

H1: μ > Δ; p-value is the area to the right of z

H1: μ < Δ; p-value is the area to the left of z

H1: μ ≠ Δ; p-value is the area in the tails greater than |z|

If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true.

If the p-value is greater than the significance level, i.e., p-value > α, then we fail to reject the null hypothesis and conclude that the null is plausible.  Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible.

The hypothesis test in this question is:

H0: μ ≥ 38 vs. H1: μ < 38

The test statistic is:

z = ( 37.5 - 38 ) / ( 1.2 / √ ( 47 ))

z = -2.856523

The p-value = P( Z < z )

= P( Z < -2.856523 )

=  0.002141547

Since the p-value is less than the significance level we reject the null hypothesis and conclude the alternate hypothesis μ < 38 is true.