n a sample of 150 Planter’s Mixed Nuts, 15 were found to be almonds. (a) Construct a 95 percent confidence interval for the true proportion of almonds. (Use a z-value taken to three decimal places in your calculations. Round your answers to 3 decimal places.) The 95% confidence interval is from _________ to ________ . (b) May normality be assumed? YesNo  (c) What sample size would be needed for 95 percent confidence and an error of ±0.03? Sample size     _________

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Asked Sep 29, 2019
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n a sample of 150 Planter’s Mixed Nuts, 15 were found to be almonds.

 

(a) Construct a 95 percent confidence interval for the true proportion of almonds. (Use a z-value taken to three decimal places in your calculations. Round your answers to 3 decimal places.)

 

The 95% confidence interval is from _________ to ________ .

 

(b) May normality be assumed?

 

  • Yes
  • No

 

 

(c) What sample size would be needed for 95 percent confidence and an error of ±0.03?

 

Sample size     _________       

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Expert Answer

Step 1

According to the provided details, the sample of 150 Planter’s Mixed Nuts has been taken out of which 15 are almonds.

Therefore,

Sample size (n) = 150

Sample proportion () = 0.1.

The formula to calculate the 95% confidence interval for population proportion is:

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рx (1-P) р3p£: п вl

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Step 2

(a)

The critical value for 95% confidence level can be found using the z-table. That is,

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a 1-0.95 =0.05 0.05 = 1.960

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Step 3

Substitute the values in the confi...

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px(1- p) p= p+z =0.1+1.960x0.1x(1-0.1) 150 (0.052,0.148) The confidence interval for the true proportion of almonds is between 0.052 to 0.148.

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