To solve this problem, we must first convert all of the numbers represented by words into their numeral equivalents so we can simplify. To do this, we must remember that the ones place is the first to the left of the decimal point, and the tens place is one place farther to the left. If we have 4 tens and 3 ones, this is the same as 43 (we put the digit 4 in the tens place and the digit 3 in the ones place). If we replace our words with our new numerical value, we get:
43 * 10
To solve this, we just multiply these two numbers together, which gives us:
430
Therefore, your answer is 430.
Hope this helps!
Answer: d)In repeated samples of the same size, approximately 95 percent of the intervals constructed from the samples will capture the population difference in means.
Step-by-step explanation:
Confidence interval is constructed to estimate a range of values that could possibly contain the population parameter. This could be the population mean or population proportion. A 95 percent confidence interval does not mean 95% probability. It tells how confident that we are that the confidence interval contains the population proportion. If we construct 100 of the given confidence interval, we are confident that 95% of them would contain the true population parameter. Therefore, the correct option is
d)In repeated samples of the same size, approximately 95 percent of the intervals constructed from the samples will capture the population difference in means.
Answer:
The null hypothesis is that all the different teaching methods have the same average test scores.
H0: μ1 = μ2 = μ3 = μ4 = μ5
The alternative hypothesis is that at least one of the teaching methods have a different mean.
Ha: at least one mean is different. (μ1 ≠ μi)
Step-by-step explanation:
The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.
For the case above, let μ represent the average test scores for the teaching methods:
The null hypothesis is that all the different teaching methods have the same average test scores.
H0: μ1 = μ2 = μ3 = μ4 = μ5
The alternative hypothesis is that at least one of the teaching methods have a different mean.
Ha: at least one mean is different. (μ1 ≠ μi)
