A student records the repair cost for 13 randomly selected TVs. A sample mean of $72.19 and standard deviation of $15.88 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal.

Tatiana reads 40 pages of her book every night for x number of nights. write an expression that shows the number of pages she has read

Astudent is determining the influence of different types of food on the growth rate of spiders. she feeds one group of spiders two caterpillars twice a week. the other group received similarly-sized beetles twice a week. she records the amount of time required for individuals to reach sexual maturity. her hypothesis is that the spiders feeding on caterpillars will have a faster growth rate because caterpillars have a higher protein content compared to beetles. in this experiment, what is the dependent variable? spiders assigned to different feeding groups. time required to reach sexual maturity number of prey items fed twice a week. growth rates of each group.

On a certain portion of an experiment, a statistical test result yielded a p-value of 0.21. what can you conclude? 2(0.21) = 0.42 < 0.5; the test is not statistically significant. if the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 21% of the time, so the test is not statistically significant. 0.21 > 0.05; the test is statistically significant. if the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 79% of the time, so the test is not statistically significant. p = 1 - 0.21 = 0.79 > 0.05; the test is statistically significant.