Question

A sample of 200 bulbs made by a company give a lifetime mean
of 1540 hours with a standard deviation of 42 hours. Is it likely that the sample has been drawn from a population with a mean lifetime of 1500 hours? You may use 5 percent level of significance

Answer 1

Answer:

Probability of manufacturing a defective bulb = 2% = 0.02

Probability of manufacturing a good bulb = 1 - 0.02 = 0.98

Using the formula for binomial distribution :

\sum_{n=0}^{r}_{r}^{n}\textrm{C}\cdot p^r\cdot q^{n-r}

Let X be the random variable for manufacturing the defective bulbs

a). n = 200 bulbs and r = 2

P(X < r = 2) = \sum_{n=0}^{2}_{r}^{200}\textrm{C}\cdot 0.02^r\cdot 0.98^{200-r}\\\\P(X < r = 2) = _{0}^{200}\textrm{C}\cdot 0.02^0\cdot 0.98^{200-0}+_{1}^{200}\textrm{C}\cdot 0.02^1\cdot 0.98^{199}\\\\\implies P(X < r = 2) =0.0894

b). n = 200 and r = 3

P(X < 4 ) = P(X < 2) + P(2) + P(3)

= 0.0894 + 0.34212

= 0.43

⇒ P(X > 3) = 1 - 0.43

= 0.57

Answer 2

Answer:

Computed value of Z = 13.47 lies in rejection region. Therefore null hypotheses is rejected. The average life of the bulb is significantly different from 1500hours.

Explanation:

Given: The sample size, n=200

Sample mean equals to 1540 hours

The sample standard deviation (s) = 42 hours

The null and alternative hypotheses:

              $H_{o}:\mu=1500hours\\H_{I}:\mu\neq 1500hrs$

The value of $\alpha =0.05$

      $Z=\frac{X-\mu_{H_{o}}}{\sigma_{X}}$

where $\mu_{H_{o}}$ is the value of $\mu$ that null hypotheses is true.

         $\sigma_{X}$ is estimated mean's standard error

Here, $\mu_{H_{o}}=1500$,

$\sigma_{X}=\frac{\sigma}{\sqrt{n} } =\frac{s}{\sqrt{n} }=\frac{42}{\sqrt{200} }=2.97$

Now,   $Z=\frac{X-\mu_{H_{o}}}{\frac{s}{\sqrt{n} } }$

  ⇒      $Z=\frac{1540-1500}{2.97}$

  ⇒       $Z=\frac{40}{2.97}$

 ∴         $Z=13.47$

Therefore the computed value of Z is  13.47.