Please answer d,e,f,g The life in hours of an Electronic sensor is known to be a

Question

Please answer d,e,f,g

The life in hours of an Electronic sensor is known to be approximately normally distributed, with standard deviation sigma = 20 hours. A random sample of 10 sensors resulted in the following data: 500, 550, 560, 575, 525, 505, 510, 540, 550, 545 (a) Is there evidence to support the claim that mean life exceeds 520 hours? Use a fixed-level test with alpha = 0.05. (b) What is the P-value for this test? Conclude the same of part (a)? (c) What is the Beta-value for this test if the true mean life is 535 hours? (d) What sample size would be required to ensure that Beta does not exceed 0.10 if the true mean life is 540 hours? (e) Construct a 95% one-sided lower CI on the mean life. (f) Use the CI found in part (e) to test the hypothesis. (g) Show how to get the same result for part a) and b) using Minitab

Explanation / Answer

Here we have to test

H0 : mu = 520 Vs H1 : mu > 520

where mu is population mean.

Assume alpha = level of significance = 0.05

Assume sigma = population standard deviation = 20 hours

n = 10

Here sample size is small and population standard deviation is known so we use one saple z-test.

The test statistic is,

Z = (Xbar - mu) / (sigma/sqrt(n))

where Xbar is sample mean = 536

mu = 520

Z = (536-520) / (20/sqrt(10)) = 2.53

Now we have to find P-value for the test.

P-value we can find by using EXCEL.

syntax :

=1- NORMSDIST(z)

where z is test statistic.

P-value = 0.006

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion : There is sufficient evidence to say that mean life exceeds 520 hours.

Here beta is nothing but the type II error.

Type II error = P(Accept H0 / H1)

We can find power in MINITAB.

steps :

STAT --> Power and sample size -- >One sample Z --> sample sizes : 10 --> differences : 535-520 = 15 --> Standard deviation : 20 --> Options --> Alternative hypothesis : greator than --> significance level: 0.05 --> ok --> ok

Power and Sample Size

1-Sample Z Test

Testing mean = null (versus > null)
Calculating power for mean = null + difference
Alpha = 0.05 Assumed standard deviation = 20

Sample
Difference Size Power
15 10 0.766342

Power = 0.7663

Power = 1 - beta

0.7663 = 1 - beta

beta = 1 - 0.7663 = 0.2337

Here we have to find sample size when

margin of error (E) = 0.10

sigma = 20

The sample size formula is,

n = [(Zc*sigma) / E]2

Zc is the critical value for normal distribution.

Zc we can find by using EXCEL.

syntax :

=NORMSINV(probability)

where probability = 1 - a/2

a = 1 - C

C is confidence level = 95%

Zc = 1.96

n = [(1.96*20) / 0.10]2 = 153664

MINITAB steps for one sample z-test :

ENTER data into MINITAB sheet --> Stat --> Basic statistics --> 1-Sample Z --> Samples in columns : select data column --> Standard deviation : 20 --> Perform hypothesis test --> Hypothesized mean : 520 --> Options --> COnfidence level : 95.0 --> Alternative : greator than --> ok --> ok

————— 4/8/2017 10:20:11 AM ————————————————————

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One-Sample Z: C1

Test of mu = 520 vs > 520
The assumed standard deviation = 20

95%
Lower
Variable N Mean StDev SE Mean Bound Z P
C1 10 536.000 25.033 6.325 525.597 2.53 0.006

We get the same result using MINITAB too.

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