Question

1. By hand solve the problem.

The life in hours of a thermocouple used in furnace is known to be approximately normally distributed, with standard deviation σ=15 hours . A random sample of 15 thermo-couples resulted in the following data:

193	192	221	183	180
200	207	187	190	219
193	222	189	190	197

1. Using Z₀. Is there evidence to support the claim that the mean is 190 hours (α=.05) ?
2. What is the P-value ?
3. Construct a 99% two-sided CI on the mean life.
4. Use the CI found in part (c) to test the hypothesis of part a

Answer 1

= 190, n= 15, = 197.533, = 15, =0.05

a)

Ho: = 190

Ha: 190

calculate z test statistics

z= 1.945

The z-test statistic is= 1.95

now calculate z critical value for two tailed test with α=0.05

using normal z table we get

The critical​ values are = ( -1.96, 1.96 )

Because the test statistic is less than the critical value

do not reject the null hypothesis.

Therefore there is enough evidence to support the claim that the mean is 190 hours

b)

calculate P-Value

P-Value = 2 * ( 1 - P(z < 1.95))

using normal z table we get

P(z < 1.95) = 0.9744

P-Value = 2 * ( 1 -  0.9744 )

P-Value = 0.0512

c)

formula for confidence interval is

Where Zc is the z critical alue for c= 99%

Zc= 2.58

87.557 < < 207.509

We get confidence interval as ( 87.557 , 207.509 )

d)

Therefore there is enough evidence to support the claim that the mean is 190 hours

because, 190 lies within our confidence interval 87.557 to 207.509