Question

A random sample of the number of farms (in thousands) in various states follows. Estimate the mean number of farms per state with 99% confidence. Assume o= 31.

16 48 79 44 49 3 90 80 9 57 8

4 64 33 54 95 47 50 40 109

A random sample of 45 four-year-olds attending day car centers provided a yearly tuition average $3850 and the population standard deviation of $620.

If day care center were starting up and wanted to keep tuition low, what would be a reasonable amount to charge.
     round to nearest hundred.

If the variance of the water temperature in a lake is 32 degrees, how many days should the researcher select to measure the temperature to estimate the true mean within 4 degrees with 90% confidence
The researcher needs a sample of at least ____ days (its NOT 5 days)

A random sample of 43 cars in the drive-thru of a popular fast food restaurant revealed an average bill of $19.99 per car. The population standard deviation is $6.49. Estimate the mean bill for all cars from the drive-thru with 91% confidence.
_____ < u < ____

For a random sample of 70 overweight men, the mean of the number of pounds that they were overweight was 28. The standard deviation of the population is 3.9 pounds.

a. the best point estimate of the mean is _____ pounds
b. Find the 95%, 90%, 99% confidence interval
   c. Find 80%, 85% interval
   d. Which interval is larger

Answer 1

Q1: ∑x =    979  

n =    20  

Mean , x̅ = Ʃx/n =    979/20 =    48.95

σ = 31

95% Confidence interval :  

At α = 0.05 two tailed critical value, z_c = ABS(NORM.S.INV(0.05/2)) =    1.960

Lower Bound = x̅ - z_c*σ/√n = 48.95 - 1.96 * 31/√20 =    35.364

Upper Bound = x̅ + z_c*σ/√n = 48.95 + 1.96 * 31/√20 =    62.536

35.364 < µ < 62.536

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Q2: x̅ = 3850, σ = 620, n = 45  

90% Confidence interval :  

At α = 0.1 two tailed critical value, z_c = ABS(NORM.S.INV(0.1/2)) =    1.645

Lower Bound = x̅ - z_c*σ/√n = 3850 - 1.645 * 620/√45 =    3697.98

Upper Bound = x̅ + z_c*σ/√n = 3850 + 1.645 * 620/√45 =    4002.02

3697.98 < µ < 4002.02  

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Q3: Population variance, σ² = 32

Margin of error, E =    4

Confidence interval, CL =    0.9

Significance level, α = 1-CL =    0.1

Critical value, z = NORM.S.INV(0.1/2) =    1.6449

Sample size, n = (z² * σ² / E²) = (1.6449² * 32 / 4²) = 5.41 = 6

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Q4: x̅ = 19.99, σ = 6.49, n = 43  

91% Confidence interval :  

At α = 0.09 two tailed critical value, z_c = ABS(NORM.S.INV(0.09/2)) =    1.695

Lower Bound = x̅ - z_c*σ/√n = 19.99 - 1.695 * 6.49/√43 =    18.3120

Upper Bound = x̅ + z_c*σ/√n = 19.99 + 1.695 * 6.49/√43 =    21.6680

18.312 < µ < 21.668        

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Q5: x̅ = 28, σ = 3.9, n = 70  

a) Point Estimate = 28

b) 95% Confidence interval :  
At α = 0.05 two tailed critical value, z_c = ABS(NORM.S.INV(0.05/2)) =    1.960
Lower Bound = x̅ - z_c*σ/√n = 28 - 1.96 * 3.9/√70 =    27.086
Upper Bound = x̅ + z_c*σ/√n = 28 + 1.96 * 3.9/√70 =    28.914
27.086 < µ < 28.914  

90% Confidence interval :  
At α = 0.1 two tailed critical value, z_c = ABS(NORM.S.INV(0.1/2)) =    1.645
Lower Bound = x̅ - z_c*σ/√n = 28 - 1.645 * 3.9/√70 =    27.233
Upper Bound = x̅ + z_c*σ/√n = 28 + 1.645 * 3.9/√70 =    28.767
27.233 < µ < 28.767  

99% Confidence interval :  
At α = 0.01 two tailed critical value, z_c = ABS(NORM.S.INV(0.01/2)) =    2.576
Lower Bound = x̅ - z_c*σ/√n = 28 - 2.576 * 3.9/√70 =    26.799
Upper Bound = x̅ + z_c*σ/√n = 28 + 2.576 * 3.9/√70 =    29.201
26.799 < µ < 29.201  

c) 80% Confidence interval :  
At α = 0.2 two tailed critical value, z_c = ABS(NORM.S.INV(0.2/2)) =    1.282
Lower Bound = x̅ - z_c*σ/√n = 28 - 1.282 * 3.9/√70 =    27.403
Upper Bound = x̅ + z_c*σ/√n = 28 + 1.282 * 3.9/√70 =    28.597
27.403 < µ < 28.597

85% Confidence interval :  
At α = 0.15 two tailed critical value, z_c = ABS(NORM.S.INV(0.15/2)) =    1.440
Lower Bound = x̅ - z_c*σ/√n = 28 - 1.44 * 3.9/√70 =    27.329
Upper Bound = x̅ + z_c*σ/√n = 28 + 1.44 * 3.9/√70 =    28.671
27.329 < µ < 28.671