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DIVAEYES
21.04.2020 •
Mathematics
Bags of a certain brand of tortilla chips claim to have a net weight of 14 ounces. Net weights actually vary slightly from bag to bag and are normally distributed with mean μ. A representative of a consumer advocate group wishes to see if there is any evidence that the mean net weight is less than advertised and so intends to test the hypotheses: H0:μ=16,Ha:μ<16.To do this, he selects 24 bags of this brand at random and determines the net weight of each. He finds the sample mean to be x= 15.68 and the sample standard deviation to be s = 0.64. Based on this data:
A) We would reject H0 at significance level 0.10 but not at 0.05.
B) We would reject H0 at significance level 0.05 but not at 0.025.
C) We would reject H0 at significance level 0.025 but not at 0.01.
D) We would reject H0 at significance level 0.01. Suppose we were not sure if the distribution of net weights was Normal. In which of the following circumstances would we not be safe using a t-procedure in this problem?
A) The mean and median of the data are nearly equal.
B) A histogram of the data shows moderate skewness.
C) A stemplot of the data has a large outlier.
D) The sample standard deviation is large.
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Ответ:
*The dot plot is shown in the attachment below
2
Step-by-step explanation:
Interquartile range is the difference between the upper median (Q3) and the lower median (Q1).
First, let's write out each value given in the data. Each dot represents a data point.
We have:
2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 7
=>Find the median:
Our median is the middle value. The middle value is the 6th value = 4
==>Upper median Q3) = the middle value of the set of data we have from the median to our far right.
2, 3, 3, 4, 4, |4,| 4, 5, [5], 6, 7
Our upper median = 5
==>Lower median(Q1) = the middle value of the data set we have from our median to our far left.
2, 3, [3], 4, 4, |4,| 4, 5, 5, 6, 7
Lower median = 3
==>Interquartile range = Q3 - Q1 = 5-3 = 2