Answer to Question #162639 in Statistics and Probability for Sunny

Question

The current rate for producing 5 amp fuses at Neary Electric Co. is 250 per hour. A new machine has been purchased and installed that, according to the supplier, will increase the production rate. A sample of 10 randomly selected hours from last month revealed the mean hourly production on the new machine was 256 units, with a standard deviation of 6 per hour. At 5% significance level can we nearly conclude that the new machine is faster?

Answer 1

We have that

$\mu = 250$

$n = 10$

$\bar x=256$

$s=6$

$\alpha=0.05$

$H_0:\mu = 250$

$H_a:\mu >250$

The hypothesis test is right-tailed.

Since the population standard deviation is unknown we use the t-test.

The critical value for 5% significance level and 9 df is 1.83

(degrees of freedom df = n – 1 = 10 – 1 = 9)

The critical region is t > 1.83

Test statistic:

$$t=\frac{\bar x - \mu}{\frac{s}{\sqrt n}}=\frac{256- 250}{\frac{6}{\sqrt {10}}}=3.16$$

Since 3.16 > 1.83 thus t falls in the rejection region we reject the null hypothesis.

At the 5% significance level the data do provide sufficient evidence to support the claim. We are 95% confident to conclude that that the new machine is faster.