Sea level is considered to be 0 feet. A probe released at sea level dropped at a constant rate for 3.25 minutes, reaching an ele
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Answer:
Step-by-step explanation:
The mean SAT score is , we are going to call it \mu since it's the "true" mean
The standard deviation (we are going to call it ) is
Next they draw a random sample of n=70 students, and they got a mean score (denoted by ) of
The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.
- So the Null Hypothesis
- The alternative would be then the opposite
The test statistic for this type of test takes the form
and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.
With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.
Answer:
Between 12.614 years and 16.386 years
Step-by-step explanation:
Given that:
Mean age (μ) = 14.5 years, standard deviation (σ) = 4.6 years, number o sample (n) and the confidence interval (c) = 90% = 0,9
α = 1 -c = 1 -0.9 = 0.1
The z score of is the same as the z score of 0.45 (0.5 - 0.05). This can be gotten from the probability distribution table. Therefore:
The margin of error (e) = =
The interval = μ ± e = 14.5 ± 1.886 = (12.614 ,16.386)