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ExtremeBDS [4]
3 years ago
7

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

Mathematics
1 answer:
Romashka [77]3 years ago
5 0

Answer:

Question 1

The correct option is  A

Question two

 The correct option is B

Question three

   SE= 0.5  

Question four

   z = - 3

Question five

The correct option is A

Step-by-step explanation:

   The population mean is  \mu =  50 \ grams

   The sample size is  n = 16

   The population size is N  = 4000

   The sample mean is  \= x  =  48.5 \grams

   The standard deviation is s  =  2 \ gram(s)

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

         SE= \frac{s }{\sqrt{n} }

=>      SE= \frac{2 }{\sqrt{16} }

=>       SE= 0.5    

The null hypothesis is  H_o :  \mu  =  50

The alternative  hypothesis is  H_a :  \mu  <  50

Generally the test statistics is mathematically represented as

      z = \frac{ \= x  - \mu  }{ SE }

=>    z = \frac{ 48.5 - 50 }{ 0.5}

=>   z = - 3

Generally the p-value is mathematically represented as

     p-value  = P(Z <  z )

=>  p-value  = P(Z <  - 3  )

From the z-table the

    P(Z <  - 3  ) =0.0013499

=>  p-value  = 0.0013499

From the obtained values we see that

  p-value  <  \alpha

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

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Answer:

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Step-by-step explanation:

So if you're simplifying then you need to find the greatest common factor. The GCF is 9.

8 0
3 years ago
Otis has saved $18,500 so far to buy a house. He can put this amount into account that earns 5.1% simple interest, or another wi
Troyanec [42]

Answer:

J Compound interest; $298.65

Step-by-step explanation:

Interest compounding pays interest on the interest. For the same annual rate, any amount of compounding will earn more interest.

For short time periods, the effect of compounding is not great. In general, it will be a fraction of the equivalent simple interest rate. Here, the effective multiplier for annual compounding is ...

  1.051^4 = 1.22024337

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  1.22024337 -1.204 = 0.01614337

That fraction of the $18500 principal is $298.65.

Compound interest earns $298.65 more than simple interest in this scenario.

Hope this helps, have an amazing day!

8 0
3 years ago
If data for a time series analysis are collected on an annual basis only, which pattern can be ignored? a. horizontal b. seasona
ddd [48]

If data for a time series analysis are collected on an annual basis only, then the option(b) seasonal pattern can be ignored

Time series analysis is a specific way of analyzing a sequence of data points collected over an interval of time

The seasonal pattern is refers to the seasonal characteristics of the time series data. It is the predictable pattern that repeats at a certain frequency within one year, such as weekly, monthly, quarterly, etc.

A horizontal pattern exists when the data fluctuate randomly around a constant mean over time.

A trend pattern exists when there is a long-term increase or decrease in the series.

The cyclical component of a time series refers to fluctuations around the trend, excluding the irregular component, revealing a succession of phases of expansion and contraction

Here, the data for a time series analysis are collected on an annual basis, so the seasonal pattern can be ignored

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7 0
2 years ago
Lin wants to know how many games teenagers in the United States have on their phones.
Ahat [919]
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3 years ago
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leva [86]

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So, the question is: how many milliseconds are 47 seconds?

By definition, one second is 1000 milliseconds, so 47 seconds are 47000 milliseconds.

7 0
3 years ago
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