Download Since the control group consists of a sample of strawberries, the

Since the control group consists of a sample of strawberries, the given standard deviation is the
sample standard deviation, s.
sum of squared deviations from mean
one less than the sample size
square root returns this measure to the original units of the data set
The standard deviation for a sample (or a population) may be roughly interpreted as the average, or typical, deviation (difference) from the mean of the sample (or population).
In the context of this problem, the typical, or "average," difference between the individual discoloration ratings and the mean discoloration rating of strawberries in the control group is 2.141.
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Random Assignment?
50 strawberries
Pair Share
control group (no preservative)
25 strawberries
treatment group (preservative)
25 strawberries
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The preservative appears to be effective in lowering the amount of discoloration in strawberries because the center of the distribution of discoloration ratings in the treatment group (median = 5 ) is clearly lower than the center of the distribution of ratings in the control group (median = 7).
Since 0 is not contained in the 95% confidence interval, there is significant evidence of a difference between the population mean discoloration ratings of strawberries not treated and those treated with preservative. margin of error
A statistic is an estimate of a parameter.
statistic =
95%
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(d) Conditions
(1) Simple random sample
(2) Normal distribution of x
(3) Sample size less than 10% of population size
formula sheet:
statistic ± critical value standard deviation of statistic
90%
t
­1.71
1.71
[4.35, 5.97]
Interpreting the interval...
I am 90% confident that the population mean discoloration rating for strawberries treated with preservative is between 4.35 and 5.97.
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(e) Conditions
(1) Simple random sample
(2) Normal distribution of x
(3) Sample size less than 10% of population size
99%
­2.80
2.80
t
[5.40, 7.80]
Interpreting the interval...
I am 99% confident that the population mean discoloration rating for strawberries not treated with preservative is between 5.4 and 7.8.
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(f)
control
treatment
By hand...
degrees
of
freedom
[0.16, 2.72)
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(g) This is a two­sample t­test for the difference in population means.
Conditions
(1) Treatments were randomly assigned to strawberries.
(2) The distribution of discoloration ratings for both treatment and control indicate that it is reasonable to assume that the respective populations of discoloration ratings are approximately normal.
(3) There are potentially an infinite number of strawberries, so
the samples sizes of 25 are certainly less than 10% of the population sizes.
Let μ1 be the mean discoloration rating of all strawberries not treated with preservative.
Let μ2 be the mean discoloration rating of all strawberries treated with preservative.
H0: μ1 = μ2
Ha: μ1 > μ2
Test Statistic
(with 47.56 degrees of freedom)
.0142
t
2.26
P­value = Since the P­value is less than α = .05, I reject the null hypothesis. There is statistically significant evidence that the population mean discoloration rating for untreated strawberries is greater than the population mean discoloration rating for strawberries treated with preservative.
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(h) Conditions
(1) We have a random sample of strawberries treated with preservative.
(2) The distribution of discoloration ratings for treated strawberries is
mound­shaped and symmetric. This indicates that it is reasonable to
assume that the distribution of discoloration ratings for the population of
treated strawberries is approximately normal.
(3) There are certainly more than 250 strawberries in the population.
Let μ be the mean discoloration rating of all strawberries treated with preservative.
Test Statistic
(with 24 degrees of freedom)
.0439
t
­1.78
P­value = Since the P­value is less than α = .05, I reject the null hypothesis. There is statistically significant evidence that the mean discoloration rating for the population of strawberries treated with preservative is less than 6.
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Conditions
(1) (Random sample of) subjects to which each treatment
was assigned in a random order.
(2) A sample of size 46 is large enough to ensure that the
sample mean difference in times is approximately normal.**see below
(3) Population must be at least 460.
[­8.1, ­5.1]
I am 95% confident that the population mean difference in melting times between chocolate chips and
chocolate­flavored chips is in the interval ­8.1, and ­5.1. Since 0 is not contained in this interval, there is significant evidence of a difference in the mean amount of time it takes to melt the chips ­ flavored chips
take longer to melt on average than do chocolate ones.
** According to the Central Limit Theorem, a sample size of n = 46 is "large enough" to ensure that the sampling distribution of the sample mean difference in melting times will be approximately normal. Additionally, the graphs below support the assertion that
the population of differences is itself approximately normal.
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(4) I will conduct a matched pairs significance test for a mean. Conditions
(1) We are told that we have a random sample of 10 beef specimens.
(2) Based on the boxplot of the differences below, it is reasonable to assume that the differences come from a normally distributed population of differences.
(3) There is potentially an infinite supply of beef in the population so our sample size of 10 is certainly less than 10% of the population size.
0
sample differences (A ­ B)
Let μd be the population mean difference (A ­ B) in the amount of E. coli detected by the two methods.
Test Statistic
(with 24 degrees of freedom)
.0786
.0786
t
­1.46
1.46
P­value = Since the P­value is greater than α = .05, I fail to reject the null hypothesis. There is no significant
evidence to indicate a mean difference in the amount of E. coli detected by the two methods.
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