A system of equations is shown below:
1 answer:
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Answer:
There is strong evidence that less than 87% of the orders are delivered in less than 10 minutes.
Decision rule: Reject the null hypothesis if (P-value < level of significance)
Test statistic z=-2.70
Decision: Reject the null hypothesis (0.003<0.010)
Step-by-step explanation:
In this question we have to test an hypothesis.
The null and alternative hypothesis are:
The significance level is assumed to be 0.01.
The sample of size n=80 gives a proportion of p=61/80=0.7625.
The standard deviation is:
The statistic z is then
The P-value is
The P-value (0.003) is smaller than the significance level (0.010), so the effect is significant. The null hypothesis is rejected.
There is strong evidence that less than 87% of the orders are delivered in less than 10 minutes.
Decision rule: Reject the null hypothesis if (P-value < level of significance)
Test statistic z=-2.70
Decision: Reject the null hypothesis (0.003<0.010)
Answer:
Step-by-step explanation:
The missing image is attached below.
From the given information
The number of players is 44.
From the box plot;
The minimum weight = 154 pounds
The first quartile Q₁ = 159 which is 25% of the data below 159 pounds.
The second quartile Q₂ = 213 which is 50% of the data below 213 pounds
The third quartile Q₃ = 253 which is 75% of the data below 253 pounds
The maximum weight = 268 pounds.
Here, the value of 213 is the middle value and which signifies the median.
The 50% of the data are located both on the left side and the right side of the median.
Thus, the percentage of players weighting greater than or equal to 213 pounds is 50%.
