-41. The sodium content of tweny-five 300-gram boxes of anic cornflakes was dete

Question

-41. The sodium content of tweny-five 300-gram boxes of anic cornflakes was determined. The data (in milligrams) as follows: 131.15, 13069, 13091, 12954, 129.64, org are 28.77, 130.72. 128.33, 12824, 12965, 130.14, 129.29. 28.71. 12900. 12939, 130.42. 129.53. 130.12. 129.78, 130.92, 130.8, 129.73, 133.15, 128.77, 129.74 a) Can you support a claim that mean sodium content of this brand of cornflakes differs from 130 mg? Use -0.05 Find the P-value. (by Check that sodium content is normally distributed. (a) What sample size would he required to detect a true mean e) Explain how the question in pan (a) could be answered by c) Compute the power of the test if the true mean sodiun content is 130.5 m sodium content of·1301 mg if you wanted the power of th test to be at least 0.75m constructing a two-sided contidence inerval on the mean sodium content. 7. 9-82

Explanation / Answer

9 - 61 H0 : = 130 mg

Ha : 130 mg  

sample mean x = 129.8852

sample standard deviation s = 1.0755

standard error of sample mean sodium content se0 = s / sqrt(n) = 1.0755/ sqrt(25) = 0.2151

test statistic

t = (129.8852 - 130)/ (1.0755/ sqrt(25)) = -0.1148/ 0.2151 = -0.534

here dF+ 24 and alpha = 0.05

t0.05, 24 = 2.0639

p - value = 0.5983 < 0.05

here t < tcritical so we cannot reject the null hypothesis and can conclude that brand of cornflkes doesn't differs from 130 mg.

(b) as Variation between data values are not that much we can assume that the sodium content is normally distriubted.

(c) True mean sodium content = 130.5 mg

sample mean above we shall reject the null hypothesis = H +- t24,0.05 se0 = 130 + 2.0639 * 0.2151 = 130.444

so we shall reject the null hypothesis is the sample mean content will be above 130.444

so Pr(Type II error) = Pr(x > 130.444 ; 130.5 ; 0.2151)

Z = (130.444 - 130.5)/ 0.2151 = -0.26

so Pr(Type II error) = Pr(Z < -0.26) = 0.3974

Power = 0.6026

(d) Power of the test = 0.75

Pr(Type II error) = 0.25

let say size is n

standard error of sample mean = 1.0755/ n

so sample mean is greater than x = 130 + tdf, 0.05 (1.0755/ n )

so Power of the test = Pr(x > 130 + tdf, 0.05 (1.0755/ n ) ; 130.1 ; (1.0755/ n ) ) = 0.75

Pr(x < 130 + tdf, 0.05 (1.0755/ n ) ; 130.1 ; (1.0755/ n ) ) = 0.25

Z = -0.6745 = (130 + tdf, 0.05 (1.0755/ n ) - 130.1)/ (1.0755/ n)

-0.725/n = -0.1 + tdf, 0.05 (1.0755/ n )

here we will assume that  tdf, 0.05 = Z0.05 = 1.96

-0.725/n = -0.1 + 2.108/n

2.833/n = 0.1

n = 28.33

n = 802.5889 = 803

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