A random sample of six steel beams has mean compressive strength of 58.392 psi (pounds per = 648 psi. Test the null hypothesis H-u = 58,000 psi against the alternative hypothesis H,: H> 58,000 psi at 5% level of significance square inch) with a standard deviation of s %3D (value for t at 5 degree of freedom and 5% significance level is 2.0157). Here u denotes the population mean. (A.M.I.E., Summer 2000)
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Q: EXERCISE 1. A random sample of six steel beams has mean compressive strength of 58.392 psi (pounds…
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- A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with ² = 1000 (psi)². A random sample of 12 specimens has a mean compressive strength of x= 3250 psi. What must be the P-value in testing the hypothesis that u = 3225 psi against the alternative hypothesis, > 3225 psi?1:3۱ 1 م 150' Done T TEST EXERCISE_c4bf49f49d24... 1 من 3 EXERCISE 1. A random sample of six steel beams has mean compressive strength of 58.392 psi (pounds per square inch) with a standard deviation of s - 648 psi. Test the null hypothesis H =u = 58,000 psi against the alternative hypothesis H,: H> 58,000 psi at 5% level of significance (value for t at 5 degree of freedom and 5% significance level is 2.0157). Here u denotes the population mean. (A.M.IE., Summer 2000) 2. A certain cubical the was thrown 96 times and shows 2 upwards 184 times. Is the the biased? Ans. die is biased. 3. In a sample of 100 residents of a colony 60 are found to be wheat eaters and 40 rice eaters. Can we assume that both food articles are equally popular? 4. Out of 400 children, 150 are found to be under weight. Assuming the conditions of simple sampling, estimate the percentage of children who are underweight in, and assign limits within which the per-centage probably lies. Ans. 37.5% approx. Limits - 37.5…2 methods are used for measuring particulate matter (PM 2.5): a PMI filter based 24 hour average and a pDR-1500 direct reading 24 hour average. The PMI method has a mean of 22 ug/m^3 and a standard deviation of 2.5 ug/m3. The pDR-1500 has a mean of 18 ug/m^3 and a standard deviation of 2 ug/m^3 . You get one measurement from each device (not on the same day). What are the mean and standard deviation of the difference in PM2.5 values between the two methods (PMI – pDR-1500)? Assume that the difference in concentrations has a Normal distribution. What is the probability that the concentration you measure on the PMI is higher than on the pDR-1500?
- The Null would be accepted is that correct? 6. State the null hypothesis for variation in central-to-peripheral fat ratio (kg) from baseline to 12 months for the TCu 380A IUD group. Should the null hypothesis be accepted or rejected?One method for straightening wire prior to coiling it to make a 6. (Hypothesis Test, 2 spring is called "roller straightening". Suppose that a sample of 30 wires is selected and each is tested to determine tensile strength (N/mm²). The resulting sample mean and sample standard deviation are 2175 and 35, respectively. It is known that the mean tensile strength for spring made using spinner straightening is 2148 N/mm². (1) What is the random variable X in this problem? What does the mean µ of X represent? (2) What null hypothesis and alternative hypothesis should be tested in order to determine if the mean tensile strength for the roller method is better than the mean tensile strength for spinner method? (3) Is this one-tailed or two-tailed test? (4) What test statistic should be used to test the hypotheses? Is a normality assumption of the population necessary? Why? (5) At the significance level a = 0.05, compute the rejection region (RR). (6) Compute the value of your test statistic…The variability of sizes of the batch of screws were studied by the company owner. They took a sample of 20 screw and found out that the deviation, s ^ 2 = 0.0153m * m ^ 2 . The variation should only to be under ±0.1 mm2 to prevent reworks. At 5% CI, conduct hypothesis testing.
- A survey organization sampled 64 households in a community and found that the mean number of TV sets per household was 3.38. The standard deviation is 1.2. Can you conclude that the mean number of TV sets per household is greater than 3? Use the a = 0.01 level. a. State the null and alternative hypotheses. ^ || :°H H1 : p v b. Find the critical value and sketch the rejection region. Sketch of Rejection Region: तामोङ्गान् नाङ्गर + + + + + + + + + + 3 -3 -2 -1 0 3 -3 -2 -1 0 2 3 -3 -2 -1 0 Critical value(s): = v c. Find the test statistic. (If there are two critical values, list both values separated by a comma) d. Find the p-value. + +A machine packages shirt in bags with a mean weight of 3kg. A random sample of 20 bags has a mean weight of 3.05kg with a standard deviation of 0.05kg, from the data given, state , state the value of a parameter and a statistic.A research team measured tidal volume in 15 experimental animals , the mean and standard deviation were 45 and 5 cc, respectively. Do these data provide sufficient evidence t indicate that the population mean is greater than 40 cc? Let α = 0.05 *with steps please.
- A research team measured tidal volume in 15 experimental animals, the mean and standard deviation were 45 and 5 cc, respectively. Do these data provide sufficient evidence t indicate that the population mean is greater than 40 cc? Let a(alpha)= 0.05 *with steps and explains and equations please??A survey organization sampled 64 households in a community and found that the mean number of TV sets per household was 3.38. The standard deviation is 1.2. Can you conclude that the mean number of TV sets per household is greater than 3? Use the a = 0.01 level. a. State the null and alternative hypotheses. ^d | : °H H1 : p v b. Find the critical value and sketch the rejection region. Sketch of Rejection Region: Critical value(s): जान्न गरिन्छ जान्न + + + -3 -2 -1 1 -1 0 1 3 -3 -2 -1 1 0.01 c. Find the test statistic. 2.387 x (If there are two critical values, list both values separated by a comma) d. Find the p-value. 0.0070 3 3 X o ४A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 763 hours. A random sample of 22 light bulbs has a mean life of 736 hours. Assume the population is normally distributed and the population standard deviation is 59 hours. At =D0.05, do you have enough evidence to reject the manufacturer's claim? Cormplete parts (a) through (e). (a) Identify the ull hypothesis and alternative hypothesis. O A. Ho: u> 763 O B. H: =736 O C. Ho: us736 Ha: us 763 (claim) Ha Hz736 (claim) Ha:> 736 (claim) O D. Ho: H#763(claim) O E. Ho: u<736 (claim) F. Ho u2763 (claim) Ha: H= 763 Hai H2736 HaiH<763 (b) Identify the critical value(s). Use technology. Z0 = (Use a comma to separate answers as needed. Round to two decimal places as needed.)