Oz; MD = 14.8 Oz; Mo = Essay

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oz; MD = 14.8 oz; Mo = 14.8 oz (sum 446.1/30 cases) Sum of squares = 8.783

SD = 0.55032 oz (Variance = .302862 oz)

SE of the mean = .55 (rounded off) / Square root of 30-1= .102132436 ( or .102)

95% CI = 14.87 +/- 1.96 (.102 [rounded off]) = 15.07 to 14.67 (1.96 *.102 = .19992 or .2 rounded off) .

Testing if the sample mean is significantly lower than 16 oz. We use a one sample T test because the population mean and variance are known (e.g., 16 oz with a variance of zero; Runyon, Coleman, & Pittenger, 2000).

M > 16 oz

H1: M < 16 oz

The current sample mean is less than the population mean of 16 oz

Alpha = .05 T. critical for 29 degrees of freedom is 1.699 (

Test:

One sample T test- because pop mean is 16 with no variance

Df = 29

So then T =

14.87 -- 16 = -1.13/.1 = -11.3

.55/Square root of 30

Formally stated: t (29) = -11.3, p < .05

Explanation:

The obtained critical (absolute value) is greater than the critical value.

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Therefore the null hypothesis is rejected and we assume that the sample mean is significantly lower than the population mean of 16 oz. (Same conclusion if the two were equal: Runyon et al., 2000).

4. In the current sample we find that the mean number of ounces of soda was significantly less than the population or target number of 16 ounces. There are several possibilities:

a. The current study looked at a conglomeration of samples from all shifts in the plant. While this is a random sample, it may not really address the problem. About one third of the bottles had 15 or more ounces of soda. There may be one or two particular shifts that are shorting customers. One follow-up investigation here would be to collect data from each shift and test the mean values from each shift against each other to find out if one shift or two shifts are filling bottles significantly lower than another shift. From that point it would be important is to.....

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