What additional information can be used to prove AVUW = AYXZ by SAS?

According to a research study, parents with young children slept 6.4 hours each night last year, on average. a random sample of 18 parents with young children was surveyed and the mean amount of time per night each parent slept was 6.8. this data has a sample standard deviation of 0.9. (assume that the scores are normally distributed.) researchers conduct a one-mean hypothesis at the 5% significance level, to test if the mean amount of time parents with young children sleep per night is greater than the mean amount of time last year. the null and alternative hypotheses are h0: μ=6.4 and ha: μ> 6.4, which is a right-tailed test. the test statistic is determined to be t0=1.89 using the partial t-table below, determine the critical value(s). if there is only one critical value, leave the second answer box blank.