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eduard
2 years ago
10

A 99% confidence interval for the sugar content (per 5 oz. serving) of a shipment of strawberries is 5.9 to 8.3 grams.

Mathematics
1 answer:
gavmur [86]2 years ago
5 0

Answer:

Sample mean = 7.1

Margin of error = 0.465

Step-by-step explanation:

Formula for confidence interval is;

CI = x¯ ± zE

Where;

x¯ is sample mean

z is critical value at confidence level

E is margin of error.

z for 99% Cl is 2.58

We are told the CI is 5.9 to 8.3.

Thus;

5.9 = x¯ - 2.58E - - - - (1)

8.3 = x¯ + 2.58E - - - - (2)

Add both equations together to get;

14.2 = 2x¯

x¯ = 14.2/2

x¯ = 7.1

Put 7.1 for x¯ in eq 1 to get;

5.9 = 7.1 - 2.58E

7.1 - 5.9 = 2.58E

E = 1.2/2.58

E = 0.465

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Answer:

z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374  

p_v =P(Z  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .  

Step-by-step explanation:

1) Data given and notation

n=830 represent the random sample taken

X=282 represent the people that regularly watch the TV station’s news program

\hat p=\frac{282}{830}=0.340 estimated proportion of people that regularly watch the TV station’s news program

p_o=0.38 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.38:  

Null hypothesis:p\geq 0.38  

Alternative hypothesis:p < 0.38  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .  

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