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Oz; MD = 14.8 Oz; Mo =

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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...

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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. 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 corrections for the shift in question. b. If we look at the measures of central tendency we can see that the distribution from the sample is relatively normal (Hughes & Hase, 2010).

This indicates that for some reason the model soda being dispersed is on average a little over an ounce less than the target. The range is also an indicator of the variance in the sample (Weiss & Weiss, 2012) and the range is from 14 ounces to 16.1 ounces. What this means is that not only are the bottles receiving significantly less soda than the target is 16 ounces, but very few bottles are getting the proper amount of soda.

A quick frequency distribution of the data indicates that in fact only two bottles receives 16 ounces or more in the sample and two thirds of the sample had bottles were filled with less than 15 ounces of soda. It is doubtful that some type of random glitch would result in such a distribution and it could well be that there is some timing mechanism I the soda delivery has been mis-calibrated that results in most of.

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