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A candy factory that produces chocolate bars claims each bar weighs 50 grams, at least that is what is printed on the label. Of course, there is bound to be a little variation. An inspector randomly chooses 16 bars from one day's output of 4000 bars. The average weight of the 16 bars is only 48.5 grams. The inspector wishes to test the null hypothesis that the factory is doing what it is supposed to on this day against the alternative that the company is cheating the consumer. Assume that the weights of all the candy bars that are produced follow the normal curve with a SD of 2 gram(s).
Which significance test should be used?
A z test because the SD of the population is known and the weights follow the normal curve. You are correct. Computer's answer now shown above Previous Tries Your receipt no. is 159-2320
Do you need to use SD+ o estimate the SD of the population?
No, we're given the SD of the population, we don't need to estimate it. Incorrect.
Computer's answer now shown above ries 1/1 Previous Tries
Compute SE. (Give your answer as a decimal to two places) Submit Answer Tries 0/5
Compute the test statistic. (Round your answer to two decimals, using previously rounded answer.) Submit Answer Tries 0/5
What do you conclude?
A. Reject the null, it's very unlikely that the factory is making the candy bars 50 grams as they claim.
B. Cannot reject the null, it's plausible that the factory is making the bars 50 grams as they claim to be doing. Submit Answer Tries 0/1

Respuesta :

okpalawalter8

Answer:

Question 1

The correct option is  A

Question two

 The correct option is B

Question three

   [tex]SE= 0.5[/tex]  

Question four

   [tex]z = - 3 [/tex]

Question five

The correct option is A

Step-by-step explanation:

   The population mean is  [tex]\mu =  50 \ grams[/tex]

   The sample size is  n = 16

   The population size is N  = 4000

   The sample mean is  [tex]\= x  =  48.5 \grams[/tex]

   The standard deviation is [tex]s  =  2 \ gram(s)[/tex]

Considering the first question

 The correct option is  A

z test because the SD of the population is known and the weights follow the normal curve

Considering  the second question

The correct option is  B

Yes, because with only 20 observations the sample SD is not a great estimate of the true population SD.

Considering  the third question  

   Generally the standard error is mathematically represented as

         [tex]SE= \frac{s }{\sqrt{n} }[/tex]

=>      [tex]SE= \frac{2 }{\sqrt{16} }[/tex]

=>       [tex]SE= 0.5[/tex]    

The null hypothesis is  [tex]H_o :  \mu  =  50[/tex]

The alternative  hypothesis is  [tex]H_a :  \mu  <  50[/tex]

Generally the test statistics is mathematically represented as

      [tex]z = \frac{ \= x  - \mu  }{ SE }[/tex]

=>    [tex]z = \frac{ 48.5 - 50 }{ 0.5}[/tex]

=>   [tex]z = - 3 [/tex]

Generally the p-value is mathematically represented as

     [tex]p-value  = P(Z <  z )[/tex]

=>  [tex]p-value  = P(Z <  - 3  )[/tex]

From the z-table the

    [tex]P(Z <  - 3  ) =0.0013499[/tex]

=>  [tex]p-value  = 0.0013499[/tex]

From the obtained values we see that

  [tex]p-value  <  \alpha[/tex]

So we reject the null hypotheses

The conclusion

it's very unlikely that the factory is making the candy bars 50 grams as they claim

Ver imagen okpalawalter8

Testing the hypothesis, we have that:

  • The standard error is of 0.5.
  • The test statistic is z = -3.
  • A. Reject the null, it's very unlikely that the factory is making the candy bars 50 grams as they claim.

At the null hypothesis, we test if the factory is doing what it is supposed to, that is, the mean is of 50 grams, hence:

[tex]H_0: \mu = 50[/tex]

At the alternative hypothesis, we test if the factory is cheating the customer, that is, the mean is of less than 50 grams, hence:

[tex]H_1: \mu < 50[/tex]

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

The parameters are:

  • [tex]\overline{x}[/tex] is the sample mean.
  • [tex]\mu[/tex] is the value tested at the null hypothesis.
  • [tex]\sigma[/tex] is the standard deviation of the sample.
  • n is the sample size.

For this problem, the values of the parameters are: [tex]\overline{x} = 48.5, \mu = 50, \sigma = 2, n = 16[/tex]

Hence, the standard error is:

[tex]S_e = \frac{2}{\sqrt{16}} = 0.5[/tex]

The value of the test statistic is:

[tex]z = \frac{\overline{x} - \mu}{S_e}[/tex]

[tex]z = \frac{48.5 - 50}{0.5}[/tex]

[tex]z = -3[/tex]

The critical value for a left-tailed test, as we are testing if the mean is less than a value, with a significance level of 0.05, is of [tex]z^{\ast} = -1.645[/tex]

Since the test statistic is less than the critical value for the left-tailed test, we can reject the null hypothesis.

A similar problem is given at https://brainly.com/question/14983950

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