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Allisa [31]
3 years ago
6

Consider the following problem: The box fill weight of Frosted Flakes breakfast cereal follows a normal distribution with a mean

of 9.75 ounces and a standard deviation of 0.27 ounces. A sample of 25 boxes filled this morning showed a mean of 9.85 ounces. At the 0.05 significance level, can we conclude that the mean weight is more than 9.75 ounces per box
Mathematics
1 answer:
insens350 [35]3 years ago
3 0

Answer:

Yes, we can

Step-by-step explanation:

H0 : μ = 9.75

H1 : μ > 9.75

n = 25 ; xbar = 9.85 ; σ = 0.27

The test statistic:

T = (xbar - μ) ÷ (σ / √n)

T = (9.85 -9.75) ÷ (0.27 ÷ √25)

T = 0.1 ÷ 0.054

T = 1.85

We can obtain the Pvalue from the test statistic using the Pvalue calculator :

Pvalue at t score = 1.85 ; 24 = 0.038

If Pvalue < α ; Reject H0

α = 0.05

Since Pvalue < α ; We reject H0 and conclude that mean weight is more than 9.75 ounces per box

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Answer:

a) The 99% confidence interval would be given by (24.409;24.979)  

b) n=464

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Part a

The confidence interval for the mean is given by the following formula:  

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In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

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Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,46)".And we see that t_{\alpha/2}=2.01  

Now we have everything in order to replace into formula (1):  

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So on this case the 95% confidence interval would be given by (503.01;546.99)

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The margin of error is given by this formula:  

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And on this case we have that ME =7 msec, we are interested in order to find the value of n, if we solve n from equation (1) we got:  

n=(\frac{t_{\alpha/2} s}{ME})^2 (2)  

The critical value for 95% of confidence interval is provided, t_{\alpha/2}=2.01 from part a, replacing into formula (2) we got:  

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Answer:

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