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m_a_m_a [10]
3 years ago
5

The​ quality-control manager at a compact fluorescent light bulb​ (CFL) factory needs to determine whether the mean life of a la

rge shipment of CFLs is equal to 7450 hours. The population standard deviation is 1 comma 350 hours. A random sample of 81 light bulbs indicates a sample mean life of 7 comma 240 hours. a. At the 0.05 level of​ significance, is there evidence that the mean life is different from 7450 hours question mark b. Compute the​ p-value and interpret its meaning. c. Construct a 95​% confidence interval estimate of the population mean life of the light bulbs. d. Compare the results of​ (a) and​ (c). What conclusions do you​ reach
Mathematics
1 answer:
stira [4]3 years ago
8 0
<h2>Answer with explanation:-</h2>

Let \mu be the population mean .

By observing the given information, we have :-

H_0:\mu=7450\\\\H_a:\mu\neq7450

Since the alternative hypotheses is two tailed so the test is a two-tailed test.

We assume that the  life of a large shipment of CFLs is normally distributed.

(a) Given : Sample size :  n=81 , since n>30 so we use z-test.

Sample mean : \overline{x}=7240

Standard deviation : \sigma=1350

Test statistic for population mean :-

z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

\Rightarrow\ z=\dfrac{7240-7450}{\dfrac{1350}{\sqrt{81}}}\\\\\Rightarrow\ z=-1.4

The critical value (two-tailed) corresponds to the given significance level :-

z_{\alpha/2}=z_{0.025}=1.96

Since the observed value of z (-1.4) is less than the critical value (1.96) , so we do not reject the null hypothesis.

Hence, we conclude that we have enough evidence to accept that the mean life is different from 7450 hours .

(b) The p-value : 2P(z>-1.4)=0.1615 , it means that the probability that the life of CFLs less than 7240 and greater than 7240 is 0.1615.

(c) The confidence interval for population mean is given by :-

\overline{x} \pm\ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=7240\pm(1.96)\dfrac{1350}{\sqrt{81}}\\\\=7240\pm294\\\\=(6946,7534)

,

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zysi [14]

Answer:

a) $520

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c) Interest amount is same each year

Step-by-step explanation:

Given - Georgie put $500 in her savings account, earning interest at a rate of 4% each year. She did not make any more deposits or withdrawals.

To find - a) How much money was in the account after one year?

              b) How much money was in the account after 4 years?

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Proof -

Here given that,

Principal amount = $500

rate of interest = 4% = 4/100 = 0.04

Now,

a)

Amount =  P [ 1 + RT ]

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             = 500 [ 1 + 0.04] = 520

⇒Amount = $520

b)

Amount =  P [ 1 + RT ]

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c)

In 2nd year,

Amount =  P [ 1 + RT ]

             = 500 [ 1 + 0.04(2)]

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Now,

Interest in 1st year = 520 - 500 = 20

Interest in 2nd year = 540 - 520 = 20

So,

The interest amount  is same each year

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Answer:0.565

Step-by-step explanation:

Hat contain 100 coins out of which 99 is fair and one is double headed.

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P(Double head |7 heads )=\frac{0.01\times 1}{0.99\times 0.5^7+0.01\times 1}

=\frac{0.01}{0.99\times 0.0078+0.01}=0.565

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